A family of functions can fail to converge for two very different reasons. It may oscillate too much, or it may escape vertically. Uniform boundedness isolates the second obstruction. Before asking whether a sequence of functions has a pointwise limit, a uniformly convergent subsequence, or a convergent integral, we first ask whether all functions in the sequence live inside one fixed vertical strip. The word "uniformly" matters because ordinary boundedness of each function is too weak. A sequence may consist entirely of bounded functions while the bounds needed for the individual functions grow without limit. In that case there is no single number controlling the whole sequence, and many compactness and convergence arguments lose their first estimate. [example: Bounded Terms Without a Uniform Bound] Let $E=[0,1]$, and for each $n \in \mathbb N$ define \begin{align*} f_n(x)=nx,\qquad x \in E. \end{align*} For a fixed $n$, every $x \in [0,1]$ satisfies $0 \le x \le 1$. Multiplying this inequality by the positive number $n$ gives \begin{align*} 0 \le nx \le n. \end{align*} Since $f_n(x)=nx$, this is \begin{align*} 0 \le f_n(x) \le n. \end{align*} Thus the single function $f_n$ is bounded on $E$, with bound depending on $n$. The sequence $(f_n)$ is not uniformly bounded on $E$. To see this, let $M \ge 0$ be any proposed common bound. Choose $n \in \mathbb N$ with $n>M$, and evaluate at $x=1 \in E$. Then \begin{align*} f_n(1)=n\cdot 1=n. \end{align*} Hence \begin{align*} |f_n(1)|=|n|=n>M. \end{align*} So this $M$ does not bound all values $|f_n(x)|$ at once. The failure is exactly that the valid bound for the individual term $f_n$ grows with $n$, so there is no single constant controlling the whole sequence. [/example] This example shows why uniform boundedness is a property of the whole sequence, not a property that can be checked term by term without keeping track of the constants. It is the first layer of control in many arguments: once a common bound exists, pointwise operations, limiting procedures, dominated convergence arguments, and compactness theorems can begin to operate. ## Definition ### Common Bounds A sequence is an ordered list indexed by $\mathbb N$. In this chapter the terms are functions, so the phrase "bounded sequence" must not be confused with a bounded numerical sequence. The question is whether all the function values, for all indices and all points in the domain, are bounded by the same number. [definition: Uniformly Bounded Sequence] Let $E \subset \mathbb R$, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. The sequence $(f_n)$ is uniformly bounded on $E$ if there exists $M \ge 0$ such that \begin{align*} |f_n(x)| \le M \end{align*} for every $n \in \mathbb N$ and every $x \in E$. [/definition] The order of the quantifiers is the entire definition: one number $M$ must work simultaneously for every function and every point. If $M$ is allowed to depend on $n$, the condition collapses to the separate boundedness of each individual function. [remark: Quantifier Order] Uniform boundedness says \begin{align*} \exists M \ge 0\,\forall n \in \mathbb N\,\forall x \in E,\quad |f_n(x)| \le M. \end{align*} Separate boundedness of the terms says \begin{align*} \forall n \in \mathbb N\,\exists M_n \ge 0\,\forall x \in E,\quad |f_n(x)| \le M_n. \end{align*} The first statement is stronger because the bound is chosen before the index $n$ is known. [/remark] ### Norm Language Many arguments need to replace repeated pointwise estimates by a single numerical sequence. Without a canonical number attached to each function, the phrase "one bound works for all functions" remains tied to two variables at once. For a [bounded function](/page/Bounded%20Function), the relevant obstruction is its largest absolute size on the domain, and taking the supremum records exactly the least possible uniform bound for that function. [definition: Supremum Norm on Bounded Functions] Let $E \subset \mathbb R$ be nonempty, and let $\mathcal B(E,\mathbb R)$ denote the set of bounded functions from $E$ to $\mathbb R$. The supremum norm on bounded functions is the function $\|\cdot\|_{\infty,E}: \mathcal B(E,\mathbb R) \to [0,\infty)$ defined by \begin{align*} \|f\|_{\infty,E}:=\sup_{x \in E} |f(x)|. \end{align*} [/definition] With this notation, uniform boundedness can be read as a numerical boundedness condition on the sequence $(\|f_n\|_{\infty,E})$. This reformulation is often the most efficient way to use the definition. [quotetheorem:9522] The theorem is not just notation. It says that a question about two variables, the index $n$ and the point $x$, can be compressed into one sequence of norms. Many compactness arguments begin by proving a bound in a norm and then invoking this characterisation. [example: A Uniform Bound From a Supremum Estimate] Let $E=\mathbb R$, and for each $n \in \mathbb N$ define \begin{align*} f_n(x)=\frac{\sin(nx)}{1+x^2},\qquad x \in \mathbb R. \end{align*} Fix $n \in \mathbb N$ and $x \in \mathbb R$. The sine function satisfies $-1 \le \sin(nx) \le 1$, so $|\sin(nx)| \le 1$. Also $x^2 \ge 0$, hence \begin{align*} 1+x^2 \ge 1. \end{align*} In particular $1+x^2>0$, so \begin{align*} |f_n(x)|=\left|\frac{\sin(nx)}{1+x^2}\right|. \end{align*} Since the denominator is positive, \begin{align*} \left|\frac{\sin(nx)}{1+x^2}\right|=\frac{|\sin(nx)|}{1+x^2}. \end{align*} Using $|\sin(nx)| \le 1$ gives \begin{align*} \frac{|\sin(nx)|}{1+x^2} \le \frac{1}{1+x^2}. \end{align*} Using $1+x^2 \ge 1$ gives \begin{align*} \frac{1}{1+x^2} \le 1. \end{align*} Combining these inequalities, \begin{align*} |f_n(x)| \le 1. \end{align*} The same constant $1$ works for every $n \in \mathbb N$ and every $x \in \mathbb R$, so $(f_n)$ is uniformly bounded on $\mathbb R$ with common bound $M=1$. The estimate controls only the vertical size of the functions; it does not prevent the oscillations of $\sin(nx)$ from becoming faster as $n$ grows. [/example] The example separates vertical control from oscillatory control. Uniform boundedness does not say that nearby points have nearby values, and it does not say that the sequence converges. It says only that the sequence cannot escape to arbitrarily large heights. ## Uniform Versus Pointwise Control ### Pointwise Bounds The most common mistake is to replace a common bound by a family of pointwise bounds. For each fixed $x$, the numerical sequence $(f_n(x))$ may be bounded, yet the bounds may become worse as $x$ moves through the domain. Uniform boundedness forbids that dependence on $x$. Pointwise boundedness answers a weaker question: at each point, does the vertical sequence stay finite? This is often the first condition in compactness theorems, but it is not enough to control the functions globally. [definition: Pointwise Bounded Sequence] Let $E \subset \mathbb R$, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. The sequence $(f_n)$ is pointwise bounded on $E$ if for every $x \in E$ there exists $M_x \ge 0$ such that \begin{align*} |f_n(x)| \le M_x \end{align*} for every $n \in \mathbb N$. [/definition] Any common bound should automatically give a pointwise bound by reusing the same constant at every point. This is the first comparison one must check: if the stronger condition failed to imply the weaker one, the terminology would not match the quantifier order. The only issue is whether the same constant can be recycled after the point $x$ has been fixed. This implication is also the baseline for every later compactness argument involving bounded families of functions. Before searching for examples where pointwise bounds fail to be uniform, we need the formal one-way comparison that says exactly what information is preserved when a global estimate is restricted to a single point. [quotetheorem:8536] The reverse implication fails in a way that is important for analysis. The failure happens because the pointwise bound can move with $x$, so the sequence can form taller and taller spikes at different locations. [example: Pointwise Bounded But Not Uniformly Bounded] Let $E=(0,1]$, and define $f_n:E\to\mathbb R$ by \begin{align*} f_n(x)=n\mathbb 1_{(0,1/n]}(x). \end{align*} Here $\mathbb 1_{(0,1/n]}$ denotes the indicator function of the interval $(0,1/n]$: it has value $1$ at points of $(0,1/n]$ and value $0$ outside that interval. Fix $x\in(0,1]$. For each $n\in\mathbb N$, either $x\in(0,1/n]$ or $x\notin(0,1/n]$. In the first case, $x\le 1/n$, and multiplying by the positive number $n/x$ gives \begin{align*} n\le \frac{1}{x}. \end{align*} Hence \begin{align*} |f_n(x)|=\left|n\mathbb 1_{(0,1/n]}(x)\right|=n\le \frac{1}{x}. \end{align*} In the second case, $\mathbb 1_{(0,1/n]}(x)=0$, so \begin{align*} |f_n(x)|=|n\cdot 0|=0\le \frac{1}{x}. \end{align*} Thus, for this fixed $x$, the bound $M_x=1/x$ satisfies $|f_n(x)|\le M_x$ for every $n\in\mathbb N$. Therefore $(f_n)$ is pointwise bounded on $E$. The sequence is not uniformly bounded on $E$. Let $M\ge 0$ be any proposed common bound. Choose $n\in\mathbb N$ with $n>M$. Since $0<1/n\le 1$, the point $1/n$ belongs to $E$, and also $1/n\in(0,1/n]$. Therefore \begin{align*} f_n(1/n)=n\mathbb 1_{(0,1/n]}(1/n)=n\cdot 1=n. \end{align*} So \begin{align*} |f_n(1/n)|=|n|=n>M. \end{align*} No single constant $M$ can bound all values $|f_n(x)|$ at once; the pointwise bounds exist, but they depend on the point $x$. [/example] ### Compact Domains and Envelopes The spike example is a model failure. At every fixed point the sequence eventually settles down, but the locations of the large values move toward $0$. A useful way to recover a uniform bound is to dominate the whole sequence by one bounded envelope. On a compact domain, a continuous nonnegative envelope has a finite maximum. This converts a pointwise-looking inequality into the uniform estimate needed for the whole sequence. [quotetheorem:9523] The theorem is a useful way to convert local-looking estimates into global ones. It does not say that every [pointwise bounded sequence](/page/Pointwise%20Bounded%20Sequence) on a compact set is uniformly bounded; it requires a single continuous envelope $g$. ## Stability Under Operations ### Algebraic Combinations Once a common bound has been obtained, it should behave well under the operations used in analysis. The point is not merely that sums and products are defined; the point is that their bounds can be computed from the original bounds without inspecting each index separately. A sequence may be transformed by addition, scalar multiplication, products, or composition with a bounded scalar function. Each operation has a predictable effect on the common bound, and these estimates are often the bookkeeping behind longer convergence arguments. [quotetheorem:9524] These estimates are deliberately quantitative. If a later argument needs to dominate a whole expression, the constants $M+N$, $MN$, and $|a|M$ can be carried forward. [example: A Polynomial Expression in Uniformly Bounded Sequences] Let $(f_n)$ and $(g_n)$ be sequences of functions from $E$ to $\mathbb R$ satisfying $|f_n(x)| \le 2$ and $|g_n(x)| \le 3$ for every $n \in \mathbb N$ and every $x \in E$. Define \begin{align*} h_n(x)=f_n(x)^2-4f_n(x)g_n(x)+g_n(x). \end{align*} Fix $n \in \mathbb N$ and $x \in E$. By the triangle inequality, \begin{align*} |h_n(x)|=\left|f_n(x)^2-4f_n(x)g_n(x)+g_n(x)\right|\le |f_n(x)^2|+|-4f_n(x)g_n(x)|+|g_n(x)|. \end{align*} Using $|ab|=|a||b|$ and $|a^2|=|a|^2$ for [real numbers](/page/Real%20Numbers), \begin{align*} |f_n(x)^2|+|-4f_n(x)g_n(x)|+|g_n(x)|=|f_n(x)|^2+4|f_n(x)||g_n(x)|+|g_n(x)|. \end{align*} Since $|f_n(x)|\le 2$, squaring both sides gives \begin{align*} |f_n(x)|^2\le 2^2=4. \end{align*} Since $|f_n(x)|\le 2$ and $|g_n(x)|\le 3$, multiplying the nonnegative inequalities gives \begin{align*} 4|f_n(x)||g_n(x)|\le 4\cdot 2\cdot 3=24. \end{align*} Also, \begin{align*} |g_n(x)|\le 3. \end{align*} Combining the three estimates, \begin{align*} |h_n(x)|\le 4+24+3=31. \end{align*} The same constant $31$ works for every $n \in \mathbb N$ and every $x \in E$, so $(h_n)$ is uniformly bounded on $E$. The calculation shows how a common bound survives a fixed algebraic expression. [/example] ### Subsequences Compactness arguments repeatedly pass from a sequence to a subsequence. To keep a uniform estimate after such a passage, we need the standard notion of a subsequence in this function-valued setting. [definition: Subsequence of a Sequence of Functions] Let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. A subsequence of $(f_n)$ is a sequence $(f_{n_k})_{k=1}^{\infty}$ where $n_1<n_2<\cdots$ are natural numbers. [/definition] A subsequence cannot create larger function values than were already present in the original sequence. The next inheritance property is what allows estimates to survive extraction of subsequences. [quotetheorem:9525] The result is simple, but it is used constantly: after extracting a subsequence for convergence, the original uniform estimate remains valid. ## Uniform Limits and Passing Bounds to the Limit ### Bounds From Uniform Convergence Uniform boundedness is often obtained before convergence is known, but convergence can also produce uniform boundedness. If a sequence converges uniformly to a bounded function, then all sufficiently late terms lie in a fixed neighbourhood of the limit, and the finitely many early terms can be absorbed into one maximum. This is a typical finite-plus-tail argument. [Uniform convergence](/page/Uniform%20Convergence) controls the tail with one error bound; boundedness of finitely many early functions supplies the remaining constants. [definition: Uniform Convergence] Let $E \subset \mathbb R$, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. Let $f: E \to \mathbb R$. The sequence $(f_n)$ converges uniformly to $f$ on $E$ if for every $\varepsilon > 0$ there exists $N \in \mathbb N$ such that \begin{align*} |f_n(x)-f(x)| < \varepsilon \end{align*} for every $n \ge N$ and every $x \in E$. [/definition] The definition asks for a single index $N$ that works for every point. That same uniformity is what allows boundedness of the limit to control the whole tail of the sequence. The elementary principle needed here is a finite-plus-tail one: if $f$ is bounded, $f_n \to f$ uniformly on $E$, and the finitely many early functions before the uniform tail are bounded, then the whole sequence $(f_n)$ is uniformly bounded on $E$. Choose $N$ so that $|f_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in E$; the tail is then bounded by the bound for $f$ plus $1$, and the remaining finitely many functions are absorbed into one maximum. The boundedness of the limit cannot be removed. If the limit is unbounded, uniform convergence to it is compatible with every term being unbounded or with no useful global bound on the family. [example: Uniform Convergence Does Not Create a Bound From an Unbounded Limit] Let $E=\mathbb R$, and for each $n\in\mathbb N$ define $f_n:\mathbb R\to\mathbb R$ by $f_n(x)=x+1/n$. We show that $f_n\to f$ uniformly on $\mathbb R$, where $f(x)=x$, but that the sequence is not uniformly bounded. Fix $n\in\mathbb N$ and $x\in\mathbb R$. Then \begin{align*} f_n(x)-f(x)=\left(x+\frac{1}{n}\right)-x=\frac{1}{n}. \end{align*} Since $n\in\mathbb N$, we have $1/n>0$, so \begin{align*} |f_n(x)-f(x)|=\left|\frac{1}{n}\right|=\frac{1}{n}. \end{align*} Thus the set of all error values is \begin{align*} \left\{|f_n(x)-f(x)|:x\in\mathbb R\right\}=\left\{\frac{1}{n}\right\}. \end{align*} Therefore \begin{align*} \sup_{x\in\mathbb R}|f_n(x)-f(x)|=\sup\left\{\frac{1}{n}\right\}=\frac{1}{n}. \end{align*} Given $\varepsilon>0$, choose $N\in\mathbb N$ with $N>1/\varepsilon$. If $n\ge N$, then \begin{align*} \frac{1}{n}\le \frac{1}{N}<\varepsilon. \end{align*} Hence for every $n\ge N$ and every $x\in\mathbb R$, \begin{align*} |f_n(x)-f(x)|=\frac{1}{n}<\varepsilon, \end{align*} so $f_n\to f$ uniformly on $\mathbb R$. Now fix $n\in\mathbb N$. To see that $f_n$ is not bounded on $\mathbb R$, let $M\ge 0$ be any proposed bound for $|f_n|$. Choose \begin{align*} x=M+\frac{1}{n}+1. \end{align*} Then $x\in\mathbb R$, and \begin{align*} f_n(x)=x+\frac{1}{n}=M+\frac{1}{n}+1+\frac{1}{n}=M+1+\frac{2}{n}. \end{align*} Since $2/n>0$, \begin{align*} |f_n(x)|=M+1+\frac{2}{n}>M. \end{align*} So no $f_n$ is bounded on $\mathbb R$, and consequently the sequence cannot be uniformly bounded on $\mathbb R$. Uniform convergence controls the distance from $f_n$ to the limit $f$, but here the limit $f(x)=x$ is itself unbounded on the unbounded domain. [/example] ### Bounds Passing to Pointwise Limits When a uniformly bounded sequence already has a pointwise limit, the natural question is whether the limiting function can leave the common vertical strip. Since each fixed value is a limit of numbers lying in that strip, the same bound passes to the limit. [quotetheorem:9149] This theorem explains why uniform boundedness is a closed condition under pointwise limits. It does not produce convergence; it only says that if convergence occurs, the limiting function remains inside the same vertical strip. ## Integration and Dominating Functions ### Finite Measure Domains For integration, uniform boundedness is valuable because it can supply a dominating function. On a set of finite measure, the constant function $M$ is integrable, so a uniformly bounded sequence automatically has a common $L^p$ bound for every finite $p$. This bridge from pointwise estimates to integral estimates requires a size condition on the domain. The next definition names that condition in the one-dimensional Lebesgue setting used throughout this page. [definition: Finite Measure Domain] Let $E \subset \mathbb R$ be Lebesgue measurable. The set $E$ has finite measure if \begin{align*} \mathcal L^1(E) < \infty. \end{align*} [/definition] Finite measure is needed because a constant bound is integrable only when the domain has finite measure. The point to control is not just each value $|f_n(x)|$, but the accumulated size of $f_n$ over all of $E$. Once $|f_n|$ is dominated by the constant $M$, the remaining obstruction is the measure of the domain, which determines whether that constant domination produces a finite $L^p$ bound. [quotetheorem:9526] The finite-measure hypothesis is not cosmetic. Without it, a uniformly bounded sequence may fail to be bounded in $L^p$ because the domain is too large. [example: Uniformly Bounded But Not $L^1$-Bounded on an Infinite Domain] Let $E=\mathbb R$, and for each $n\in\mathbb N$ define $f_n:\mathbb R\to\mathbb R$ by \begin{align*} f_n(x)=\mathbb 1_{[0,n]}(x). \end{align*} For any $x\in\mathbb R$, the indicator value is either $0$ or $1$, so \begin{align*} |f_n(x)|=|\mathbb 1_{[0,n]}(x)|\le 1. \end{align*} The constant $1$ is independent of both $n$ and $x$, hence $(f_n)$ is uniformly bounded on $\mathbb R$. For the $L^1$ norm, since $\mathbb 1_{[0,n]}(x)\ge 0$ for every $x$, \begin{align*} |f_n(x)|=|\mathbb 1_{[0,n]}(x)|=\mathbb 1_{[0,n]}(x). \end{align*} Therefore, by the definition of the $L^1$ norm, \begin{align*} \|f_n\|_{L^1(\mathbb R)}=\int_{\mathbb R}|f_n(x)|\,d\mathcal L^1(x). \end{align*} Substituting the previous identity gives \begin{align*} \|f_n\|_{L^1(\mathbb R)}=\int_{\mathbb R}\mathbb 1_{[0,n]}(x)\,d\mathcal L^1(x). \end{align*} The integral of an indicator function is the [Lebesgue measure](/page/Lebesgue%20Measure) of its set, so \begin{align*} \int_{\mathbb R}\mathbb 1_{[0,n]}(x)\,d\mathcal L^1(x)=\mathcal L^1([0,n]). \end{align*} Since the interval $[0,n]$ has length $n-0=n$, \begin{align*} \|f_n\|_{L^1(\mathbb R)}=n. \end{align*} Given any proposed bound $M\ge 0$, choose $n\in\mathbb N$ with $n>M$; then $\|f_n\|_{L^1(\mathbb R)}=n>M$. Thus the sequence is uniformly bounded pointwise, but its $L^1$ norms are not bounded on the infinite-measure domain $\mathbb R$. [/example] ### Dominated Convergence The preceding example shows that uniform boundedness is not the same as integral boundedness on every domain. On a Lebesgue measurable finite-measure domain, however, the constant common bound becomes an integrable dominating function, so the [dominated convergence theorem](/theorems/4) applies. [quotetheorem:4] This theorem is often the first place students see why a uniform bound is stronger than a pointwise bound. Pointwise boundedness gives no single integrable dominating function; uniform boundedness on a finite-measure set gives one immediately. ## Compactness and Equicontinuity ### Oscillation as a Separate Obstruction Uniform boundedness alone does not give a convergent subsequence. A sequence can remain inside a fixed vertical strip while oscillating faster and faster. Compactness for functions usually needs a second condition controlling horizontal behaviour. The missing condition is equicontinuity. It prevents the functions from changing too rapidly, and together with uniform boundedness it leads to the Arzela-Ascoli theorem. [definition: Equicontinuous Sequence] Let $E \subset \mathbb R$, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. The sequence $(f_n)$ is equicontinuous on $E$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for every $n \in \mathbb N$ and all $x,y \in E$, \begin{align*} |x-y|<\delta \implies |f_n(x)-f_n(y)|<\varepsilon. \end{align*} [/definition] The same $\delta$ must work for every member of the sequence. This mirrors uniform boundedness: a single constant controls the vertical size, and a single modulus controls the horizontal variation. [example: Uniformly Bounded But Not Equicontinuous] Let $E=[0,2\pi]$, and define $f_n:E\to\mathbb R$ by $f_n(x)=\sin(nx)$. We first verify the common vertical bound. For every $n\in\mathbb N$ and every $x\in E$, the sine function satisfies $-1\le \sin(nx)\le 1$, hence \begin{align*} |f_n(x)|=|\sin(nx)|\le 1. \end{align*} The same constant $1$ works for all $n$ and all $x\in E$, so $(f_n)$ is uniformly bounded on $E$. The sequence is not equicontinuous on $E$. Take $\varepsilon=1/2$. Let $\delta>0$ be arbitrary. Choose $n\in\mathbb N$ such that \begin{align*} n>\frac{\pi}{2\delta}. \end{align*} Set \begin{align*} x_n=0,\qquad y_n=\frac{\pi}{2n}. \end{align*} Since $n\ge 1$, we have $0<\pi/(2n)\le \pi/2\le 2\pi$, so $x_n,y_n\in E$. Also, \begin{align*} |x_n-y_n|=\left|0-\frac{\pi}{2n}\right|=\frac{\pi}{2n}<\delta. \end{align*} But \begin{align*} f_n(x_n)=\sin(n\cdot 0)=\sin(0)=0. \end{align*} And \begin{align*} f_n(y_n)=\sin\left(n\cdot \frac{\pi}{2n}\right)=\sin\left(\frac{\pi}{2}\right)=1. \end{align*} Therefore \begin{align*} |f_n(y_n)-f_n(x_n)|=|1-0|=1>\frac{1}{2}. \end{align*} Thus no single $\delta$ can make all functions $f_n$ vary by less than $1/2$ on $\delta$-close points. The sequence stays inside the fixed vertical strip $[-1,1]$, but its oscillations become too rapid for equicontinuity. [/example] ### Arzela-Ascoli Compactness The oscillatory example identifies the missing hypothesis for compactness. Uniform boundedness keeps the graphs from escaping vertically, and equicontinuity prevents narrow oscillations from surviving at smaller and smaller scales. [quotetheorem:66] The theorem also explains why uniform boundedness appears as a hypothesis rather than a conclusion. It supplies the range compactness needed to extract convergent values on a dense set; equicontinuity then upgrades pointwise control to uniform convergence. [example: Uniform Boundedness and Equicontinuity Working Together] Let $E=[0,1]$, and for each $n\in\mathbb N$ define \begin{align*} f_n(x)=\frac{x}{1+nx}. \end{align*} If $x\in[0,1]$, then $x\ge 0$ and $1+nx\ge 1>0$, so \begin{align*} f_n(x)=\frac{x}{1+nx}\ge 0. \end{align*} Also, since $x\le 1$ and $n+1>0$, \begin{align*} (n+1)x\le n+1. \end{align*} Because $x\le 1$, we also have $nx\le n$, and the sharper comparison needed here is \begin{align*} (n+1)x=nx+x\le nx+1. \end{align*} Dividing by the positive number $(n+1)(1+nx)$ gives \begin{align*} \frac{x}{1+nx}\le \frac{1}{n+1}. \end{align*} In particular, \begin{align*} 0\le f_n(x)\le \frac{1}{n+1}\le 1. \end{align*} The constant $1$ works for every $n\in\mathbb N$ and every $x\in[0,1]$, so the sequence is uniformly bounded. We next verify equicontinuity. Fix $n\in\mathbb N$ and $x,y\in[0,1]$. Since $1+nx>0$ and $1+ny>0$, \begin{align*} f_n(x)-f_n(y)=\frac{x}{1+nx}-\frac{y}{1+ny}. \end{align*} Putting the fractions over the common denominator gives \begin{align*} \frac{x}{1+nx}-\frac{y}{1+ny}=\frac{x(1+ny)-y(1+nx)}{(1+nx)(1+ny)}. \end{align*} Expanding the numerator, \begin{align*} x(1+ny)-y(1+nx)=x+nxy-y-nxy=x-y. \end{align*} Therefore \begin{align*} |f_n(x)-f_n(y)|=\frac{|x-y|}{(1+nx)(1+ny)}. \end{align*} Since $x,y\ge 0$, we have $1+nx\ge 1$ and $1+ny\ge 1$, hence \begin{align*} (1+nx)(1+ny)\ge 1. \end{align*} Thus \begin{align*} |f_n(x)-f_n(y)|\le |x-y|. \end{align*} Given $\varepsilon>0$, choose $\delta=\varepsilon$. Then $|x-y|<\delta$ implies $|f_n(x)-f_n(y)|<\varepsilon$ for every $n$, so $(f_n)$ is equicontinuous on $[0,1]$. Finally, the estimate already obtained gives \begin{align*} 0\le f_n(x)\le \frac{1}{n+1} \end{align*} for every $x\in[0,1]$. Given $\varepsilon>0$, choose $N\in\mathbb N$ such that $N+1>1/\varepsilon$. If $n\ge N$, then \begin{align*} 0\le |f_n(x)-0|=f_n(x)\le \frac{1}{n+1}\le \frac{1}{N+1}<\varepsilon. \end{align*} Hence $f_n\to 0$ uniformly on $[0,1]$. This example shows the favourable compactness pattern: the functions stay in one vertical strip, vary with one common Lipschitz bound, and in this case even converge uniformly to the zero function. [/example] The example illustrates a favourable situation: uniform boundedness, equicontinuity, and an explicit formula all point toward compact behaviour. In more complex problems, Arzela-Ascoli replaces the explicit formula. ## Uniform Boundedness Principles in Function Spaces The phrase "uniformly bounded" also appears in functional analysis, where the objects being bounded are operators rather than pointwise values of functions. The underlying idea is the same: a whole sequence is controlled by one constant, not by constants depending on the index. To state this form, we move from functions on a set to bounded linear operators between normed spaces. The bound is now measured using the [operator norm](/page/Operator%20Norm). [definition: Uniformly Bounded Sequence of Operators] Let $X$ and $Y$ be normed spaces, and let $(T_n)_{n=1}^{\infty}$ be a sequence such that each $T_n: X \to Y$ is a bounded [linear map](/page/Linear%20Map). The sequence $(T_n)$ is uniformly bounded in operator norm if there exists $M \ge 0$ such that \begin{align*} \|T_n\|_{\mathcal L(X,Y)} \le M \end{align*} for every $n \in \mathbb N$. [/definition] Operator-norm boundedness controls all vectors at once, but many problems first give only vector-by-vector estimates. For a fixed vector $x$, the numbers $\|T_nx\|_Y$ may stay bounded even if the operator norms $\|T_n\|_{\mathcal L(X,Y)}$ become large on different vectors. This separates control of each orbit from control of the whole unit ball, so the weaker hypothesis needs its own precise formulation. [definition: Pointwise Bounded Sequence of Operators] Let $X$ and $Y$ be normed spaces, and let $(T_n)_{n=1}^{\infty}$ be a sequence such that each $T_n: X \to Y$ is a bounded linear map. The sequence $(T_n)$ is pointwise bounded if for every $x \in X$ there exists $M_x \ge 0$ such that \begin{align*} \|T_n x\|_Y \le M_x \end{align*} for every $n \in \mathbb N$. [/definition] A central functional-analytic question is whether vector-by-vector control can ever force one operator-norm bound for the entire sequence. Completeness of the domain space is exactly the structure that makes this upgrade possible. The quoted result is stated in a broader locally convex form than the normed-space definitions above. A [Fréchet space](/page/Fr%C3%A9chet%20Space) is a complete metrizable locally convex [topological vector space](/page/Topological%20Vector%20Space), and its topology may be described by a family of seminorms $p$. A family of maps $(T_i)_{i\in I}$ is indexed by an arbitrary set $I$, not necessarily by $\mathbb N$. For a locally convex target space $Y$, equicontinuity means uniform control of all the maps $T_i$ with respect to the seminorms defining the topologies: in the Banach-space special case, this is exactly the conclusion that the operator norms are bounded by one common constant. [quotetheorem:716] This theorem is a major warning about completeness. For ordinary sequences of functions, pointwise boundedness does not imply uniform boundedness on the domain. For bounded linear operators on a [Banach space](/page/Banach%20Space), the structure of the space prevents the same kind of moving-spike failure. This section is included as an application and comparison point: the elementary definition remains the common-bound condition, while functional analysis explains when a weaker pointwise hypothesis can force it. [example: Evaluation Functionals and the Need for Completeness] Let $X=c_{00}$ be the normed space of real sequences with finite support, equipped with \begin{align*}\|x\|_{\infty}=\sup_{k \in \mathbb N}|x_k|.\end{align*} The space $X$ is not complete. For instance, define $u^{(m)}\in c_{00}$ by $u^{(m)}_k=1/k$ for $1\le k\le m$ and $u^{(m)}_k=0$ for $k>m$. If $m>\ell$, then \begin{align*}\|u^{(m)}-u^{(\ell)}\|_{\infty}=\sup_{\ell<k\le m}\frac{1}{k}=\frac{1}{\ell+1}.\end{align*} Thus $(u^{(m)})$ is Cauchy in the supremum norm. Its pointwise limit is the sequence $u_k=1/k$, which belongs to $c_0$ because $1/k\to 0$, but $u\notin c_{00}$ because $u_k\ne 0$ for every $k\in\mathbb N$. So $c_{00}$ is not complete; its completion is $c_0$. For each $n\in\mathbb N$, define $T_n:X\to\mathbb R$ by \begin{align*}T_n(x)=n x_n.\end{align*} If $\alpha,\beta\in\mathbb R$ and $x,y\in X$, then \begin{align*}T_n(\alpha x+\beta y)=n(\alpha x_n+\beta y_n)=\alpha n x_n+\beta n y_n=\alpha T_n(x)+\beta T_n(y).\end{align*} Hence $T_n$ is linear. Also, for every $x\in X$, \begin{align*}|T_n(x)|=|n x_n|=n|x_n|\le n\sup_{k\in\mathbb N}|x_k|=n\|x\|_\infty.\end{align*} Therefore $T_n$ is bounded and $\|T_n\|_{\mathcal L(X,\mathbb R)}\le n$. To get the reverse inequality, let $e^{(n)}\in c_{00}$ be the sequence with $e^{(n)}_n=1$ and $e^{(n)}_k=0$ for $k\ne n$. Then \begin{align*}\|e^{(n)}\|_\infty=\sup_{k\in\mathbb N}|e^{(n)}_k|=1.\end{align*} Also, \begin{align*}|T_n(e^{(n)})|=|n e^{(n)}_n|=|n\cdot 1|=n.\end{align*} Since the operator norm is the supremum of $|T_n(x)|$ over all $\|x\|_\infty\le 1$, this gives $\|T_n\|_{\mathcal L(X,\mathbb R)}\ge n$. Combining both inequalities, \begin{align*}\|T_n\|_{\mathcal L(X,\mathbb R)}=n.\end{align*} Thus the operator norms are not bounded above, since for any $M\ge 0$ we may choose $n\in\mathbb N$ with $n>M$ and obtain $\|T_n\|_{\mathcal L(X,\mathbb R)}=n>M$. On the other hand, the sequence is pointwise bounded. Fix $x\in c_{00}$. Since $x$ has finite support, the set \begin{align*}S=\{k\in\mathbb N:x_k\ne 0\}\end{align*} is finite. If $S=\varnothing$, then $x_k=0$ for every $k$, so $T_n(x)=n x_n=0$ for every $n$. If $S\ne\varnothing$, set \begin{align*}M_x=\max_{k\in S} k|x_k|.\end{align*} For $n\notin S$, we have $x_n=0$, hence $|T_n(x)|=0\le M_x$. For $n\in S$, we have \begin{align*}|T_n(x)|=|n x_n|=n|x_n|\le M_x.\end{align*} Thus for this fixed $x$, the numerical sequence $(T_nx)$ is bounded. Pointwise boundedness holds, but the operator norms satisfy $\|T_n\|_{\mathcal L(X,\mathbb R)}=n\to\infty$; the example shows that the completeness hypothesis in the [uniform boundedness principle](/theorems/549) is essential. [/example] The operator theorem belongs to functional analysis, but it echoes the elementary theme of this chapter: bounds that depend on the point or vector are not the same as one bound for the whole family unless additional structure forces them to coincide. ## Beyond and Connected Topics Uniformly bounded sequences are a gateway into compactness, convergence, and functional analysis. In elementary analysis, they belong next to [sequence](/page/Sequence) concepts because they adapt numerical boundedness to sequences whose terms are functions. The key change is that each term has its own internal variable, so boundedness has two layers of quantifiers. The natural next topic is uniform convergence. Uniform boundedness is often a consequence of uniform convergence to a bounded function, and it is also preserved under pointwise limits when a limit exists. The notes [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) and [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) are good continuations for the underlying sequence and compactness ideas. Measure theory gives another direction. On finite-measure domains, uniform boundedness gives a simple dominating function, so it becomes a practical hypothesis for exchanging limits and integrals. This connects naturally with the analysis of functions developed in [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Complex analysis supplies a sharper version of the same theme. Locally uniformly bounded families of holomorphic functions are controlled enough to have compactness properties, leading toward Montel-type theorems. The analytic background is developed in [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Functional analysis turns the phrase into a theorem. The uniform boundedness principle says that pointwise boundedness of bounded linear operators on a Banach space implies uniform boundedness in operator norm. This is a structural result about complete normed spaces, not merely a quantifier manipulation. ## References Androma, [Sequence](/page/Sequence). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Walter Rudin, *Principles of Mathematical Analysis* (1976). George B. Folland, *Real Analysis: Modern Techniques and Their Applications* (1999). John B. Conway, *A Course in Functional Analysis* (1990).