A sequence of functions can behave well at every individual input while behaving badly as a family. For instance, the functions $f_k:[0,1] \to \mathbb{R}$ given by $f_k(x)=x^k$ never exceed $1$ at any fixed $x$, and for each fixed $x$ the numerical sequence $(f_k(x))_{k=1}^{\infty}$ has a finite bound. Yet the family changes shape dramatically near $x=1$, and its pointwise limit is discontinuous. Pointwise boundedness is the first vocabulary for separating these two ideas: boundedness at each input, and boundedness controlled by one constant for all inputs. This distinction matters because many compactness and convergence arguments begin with a sequence $(f_k)_{k=1}^{\infty}$ and ask what can be salvaged from partial information. Pointwise boundedness is local in the variable: each $x$ is allowed to have its own bound. It is weaker than uniform boundedness, weaker than boundedness in most function-space norms, and often too weak by itself to control continuity, integrability, or convergence. Its strength is that it is exactly the hypothesis naturally produced by pointwise estimates and exactly the conclusion naturally tested by evaluation at a point. The guiding question for this chapter is: when does controlling the numerical sequence $(f_k(x))_{k=1}^{\infty}$ separately for each $x$ give meaningful control over the original sequence of functions? The answer depends on what extra structure is present. On a bare set, pointwise boundedness is only a family of independent scalar bounds. On a compact [metric space](/page/Metric%20Space) with equicontinuity, it becomes part of Arzela-Ascoli. In a [Banach space](/page/Banach%20Space), it is the hypothesis of the [uniform boundedness principle](/theorems/549) when the functions are bounded linear operators. [example: A Bounded Sequence at Each Point with a Discontinuous Limit] Let $f_k:[0,1]\to\mathbb{R}$ be defined by $f_k(x)=x^k$. For every $x\in[0,1]$ and every $k\in\mathbb{N}$, we have $0\le x\le 1$, so $0\le x^k\le 1^k=1$. Hence \begin{align*} |f_k(x)|=|x^k|=x^k\le 1. \end{align*} Thus the sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $[0,1]$, with the valid choice $M_x=1$ for every $x\in[0,1]$. Now fix $x\in[0,1)$. Since $0\le x<1$, the geometric powers satisfy $\lim_{k\to\infty}x^k=0$, so \begin{align*} \lim_{k\to\infty}f_k(x)=\lim_{k\to\infty}x^k=0. \end{align*} At the endpoint, \begin{align*} f_k(1)=1^k=1 \end{align*} for every $k$, so \begin{align*} \lim_{k\to\infty}f_k(1)=1. \end{align*} Therefore the pointwise limit $f:[0,1]\to\mathbb{R}$ is given by $f(x)=0$ for $0\le x<1$ and $f(1)=1$. This limit is not continuous at $1$: for every $\delta>0$, choose $x\in[0,1)$ with $1-\delta<x<1$, for instance $x=1-\delta/2$ if $\delta\le 2$ and $x=1/2$ if $\delta>2$. Then $|x-1|<\delta$, but \begin{align*} |f(x)-f(1)|=|0-1|=1. \end{align*} Thus pointwise boundedness, even together with pointwise convergence, does not preserve continuity. [/example] The example also shows why the word "pointwise" must be taken seriously. The bound near $x=1$ is not supplied by a stable shape of the graphs; it is supplied separately at each point. The rest of the chapter develops this separation between pointwise information, uniform information, and compactness information. ## Definition The central property is not a new kind of function, but a way of bounding a whole sequence after an input has been fixed. The definition deliberately allows the bound to depend on the input, because this is exactly what separates pointwise information from uniform information. [definition: Pointwise Bounded Sequence] Let $X$ be a set and let $(f_k)_{k=1}^{\infty}$ be a sequence of functions $f_k:X \to \mathbb{R}$ or $f_k:X \to \mathbb{C}$. The sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $X$ if for every $x \in X$ there exists a constant $M_x < \infty$ such that \begin{align*} |f_k(x)| &\le M_x \end{align*} for every $k \in \mathbb{N}$. [/definition] The parent object behind this definition is an ordinary sequence. A sequence records a countable ordered list, and in analysis the entries may be numbers, functions, operators, sets, or other mathematical objects. [definition: Sequence] Let $A$ be a set. A sequence in $A$ is a function $a:\mathbb{N}\to A$, usually written as $(a_k)_{k=1}^{\infty}$. [/definition] To speak about pointwise boundedness, the entries cannot merely be unrelated objects: they must be functions with a shared domain, so that the same input $x$ can be evaluated across the whole list. The next definition isolates that common-domain structure before we ask whether the resulting scalar sequence is bounded at each point. [definition: Sequence of Functions] Let $X$ be a set and let $Y$ be a set. A sequence of functions from $X$ to $Y$ is a sequence $(f_k)_{k=1}^{\infty}$ such that each entry is a function $f_k:X \to Y$. [/definition] The subscript in $M_x$ is the whole point of the definition. A different point may require a different bound. In quantifier form, the definition says \begin{align*} \forall x \in X\,\exists M_x < \infty\,\forall k \in \mathbb{N}: |f_k(x)| \le M_x. \end{align*} Changing the order of the first two quantifiers gives a stronger condition, which is uniform boundedness. A reader should also notice what the definition does not require. It does not require any topology on $X$, any continuity of the functions, any convergence of the sequence, or any measurability. It is a property of evaluations and scalar bounds. ## Quantifiers and First Examples ### The Quantifier Swap The definition of pointwise boundedness has two separate choices built into it: first choose the point $x$, and only then choose a bound for the resulting numerical sequence. The most common mistake is to reverse that order and act as though one bound had already been chosen before $x$ was known. Before looking at examples, it is useful to name the stronger condition obtained by making that reversal legitimate. ### Uniform Bounds The main danger in using pointwise boundedness is losing track of which constants are allowed to depend on $x$. Many false arguments replace $M_x$ by a single $M$ without an additional compactness, continuity, or functional-analytic theorem. The next definition records the stronger condition so that the contrast is visible. [definition: Uniformly Bounded Sequence of Functions] Let $X$ be a set and let $(f_k)_{k=1}^{\infty}$ be a sequence of functions $f_k:X \to \mathbb{R}$ or $f_k:X \to \mathbb{C}$. The sequence $(f_k)_{k=1}^{\infty}$ is uniformly bounded on $X$ if there exists a constant $M < \infty$ such that \begin{align*} |f_k(x)| &\le M \end{align*} for every $x \in X$ and every $k \in \mathbb{N}$. [/definition] Uniform boundedness immediately gives pointwise boundedness by taking $M_x=M$ for every $x$. The reverse implication is false on many natural domains, and the failure is not subtle: pointwise bounds can grow without limit as the point moves through the domain. The next example isolates this quantifier reversal on a domain with no topology. [example: Pointwise Bounded but Not Uniformly Bounded] Let $X=\mathbb{N}$ and define $f_k:\mathbb{N}\to\mathbb{R}$ by \begin{align*} f_k(n)=\begin{cases} n, & k=n, \ 0, & k\ne n. \end{cases} \end{align*} Fix $n\in\mathbb{N}$. If $k=n$, then $|f_k(n)|=|n|=n$. If $k\ne n$, then $|f_k(n)|=|0|=0\le n$. Hence, for every $k\in\mathbb{N}$, \begin{align*} |f_k(n)|\le n. \end{align*} Thus the sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $\mathbb{N}$, with the point-dependent bound $M_n=n$. It is not uniformly bounded. Suppose that a uniform bound $M<\infty$ existed. Choose $n\in\mathbb{N}$ with $n>M$. Then, using the function with the same index $k=n$, \begin{align*} |f_n(n)|=|n|=n>M. \end{align*} This contradicts the required inequality $|f_k(x)|\le M$ for all $x\in\mathbb{N}$ and all $k\in\mathbb{N}$. Therefore no single finite constant bounds all values $|f_k(n)|$ at once; the example is pointwise bounded but not uniformly bounded. [/example] This example is deliberately discrete. It removes every analytic complication and leaves only the quantifiers. Pointwise boundedness says each vertical column is bounded; uniform boundedness says all vertical columns are bounded by the same ceiling. ### Moving Spikes For sequences of functions on a metric space, a pointwise bound may coexist with sharp spikes. This matters because graphs can look locally controlled at each fixed input while their sup norms diverge. The next example shows that even when every function is continuous, pointwise boundedness need not control the global height of the functions. [example: Moving Spikes] Let $f_k:[0,1]\to\mathbb{R}$ be defined by \begin{align*} f_k(x)=\begin{cases} k^2x, & 0\le x\le 1/k, \ 2k-k^2x, & 1/k<x\le 2/k, \ 0, & 2/k<x\le 1. \end{cases} \end{align*} At the joining point $x=1/k$, the left formula gives $k^2(1/k)=k$ and the middle formula gives $2k-k^2(1/k)=2k-k=k$. At the joining point $x=2/k$, when this point lies in $[0,1]$, the middle formula gives $2k-k^2(2/k)=2k-2k=0$, matching the last formula. Thus each $f_k$ is a continuous triangular spike. We show that the sequence is pointwise bounded. First, at $x=0$, \begin{align*} f_k(0)=k^2\cdot 0=0 \end{align*} for every $k$, so $|f_k(0)|\le 0$. Now fix $x\in(0,1]$. Choose an integer $N_x$ with $N_x>2/x$. If $k\ge N_x$, then $k>2/x$, so $kx>2$, hence $x>2/k$. Therefore the third branch applies and \begin{align*} f_k(x)=0 \end{align*} for every $k\ge N_x$. For the remaining finitely many indices $1\le k<N_x$, set \begin{align*} M_x=\max\{|f_1(x)|,|f_2(x)|,\ldots,|f_{N_x-1}(x)|,0\}. \end{align*} This maximum is finite because it is the maximum of finitely many [real numbers](/page/Real%20Numbers), and the previous paragraph gives $|f_k(x)|\le M_x$ for every $k\in\mathbb{N}$. Hence $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $[0,1]$. For each fixed $k$, the maximum height is exactly $k$. If $0\le x\le 1/k$, then $0\le k^2x\le k^2(1/k)=k$. If $1/k<x\le 2/k$, then $0\le 2k-k^2x<2k-k^2(1/k)=k$. If $2/k<x\le 1$, then $f_k(x)=0$. Thus $|f_k(x)|\le k$ for every $x\in[0,1]$. At $x=1/k$, \begin{align*} |f_k(1/k)|=|k^2(1/k)|=k. \end{align*} Therefore \begin{align*} \sup_{x\in[0,1]}|f_k(x)|=k. \end{align*} Since the numbers $k$ are unbounded as $k\to\infty$, no single finite constant can bound all the suprema. The sequence is pointwise bounded, but it is not uniformly bounded in the supremum norm. [/example] The spike example is a useful warning for compactness arguments. Compactness of the domain alone does not upgrade pointwise boundedness to uniform boundedness. Some additional regularity of the family, usually equicontinuity or linear structure, is required. ## Pointwise Bounds and Pointwise Convergence A common source of pointwise boundedness is convergence. If a numerical sequence converges, then it is bounded; applying this at every input gives pointwise boundedness of a pointwise convergent sequence of functions. The next definition names the convergence mode whose scalar shadows are being tested. [definition: Pointwise Convergent Sequence of Functions] Let $X$ be a set and let $(f_k)_{k=1}^{\infty}$ be a sequence of functions $f_k:X \to \mathbb{R}$ or $f_k:X \to \mathbb{C}$. The sequence $(f_k)_{k=1}^{\infty}$ converges pointwise on $X$ to a function $f:X \to \mathbb{R}$ or $f:X \to \mathbb{C}$ if for every $x\in X$, \begin{align*} \lim_{k\to\infty} f_k(x) &= f(x). \end{align*} [/definition] Once convergence has been specified point by point, a natural question is what boundedness it automatically supplies. The next theorem records the basic implication that pointwise convergence gives a finite bound along the sequence at each input, while still saying nothing about a common bound across all inputs. [quotetheorem:8528] This theorem is often used in the contrapositive direction: if at some point $x_0$ the scalar sequence $(f_k(x_0))_{k=1}^{\infty}$ is unbounded, then there is no pointwise limit at $x_0$. But boundedness alone does not force convergence. The scalar sequence may oscillate forever at each point. [example: Pointwise Bounded Without Pointwise Convergence] Let $f_k:[0,1]\to\mathbb{R}$ be the constant function $f_k(x)=(-1)^k$. For every $x\in[0,1]$ and every $k\in\mathbb{N}$, we have \begin{align*} |f_k(x)|=|(-1)^k|=1. \end{align*} Thus $|f_k(x)|\le 1$ for every $k$, so the sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded on $[0,1]$, with the valid choice $M_x=1$ for every $x\in[0,1]$. It remains to check that no pointwise limit exists. Fix $x\in[0,1]$. Along the even indices $k=2m$, \begin{align*} f_{2m}(x)=(-1)^{2m}=1. \end{align*} Along the odd indices $k=2m-1$, \begin{align*} f_{2m-1}(x)=(-1)^{2m-1}=-1. \end{align*} If the scalar sequence $(f_k(x))_{k=1}^{\infty}$ converged to some real number $L$, then its even-indexed subsequence and odd-indexed subsequence would both converge to $L$. The displayed formulas force the even subsequence to converge to $1$ and the odd subsequence to converge to $-1$, so this would require $L=1$ and $L=-1$ at the same time, impossible. Therefore $(f_k(x))_{k=1}^{\infty}$ does not converge for any $x\in[0,1]$. This example separates boundedness from convergence: the values stay inside $[-1,1]$ at every point, but they oscillate forever at every point. [/example] Boundedness becomes more powerful when paired with compactness in the scalar target. For a single point $x$, the bounded scalar sequence $(f_k(x))$ has convergent subsequences. The obstacle is choosing one subsequence that works at many points at once. ## Diagonal Selection on Countable Domains On a countable domain, pointwise boundedness does allow a global subsequence that converges at every point. The proof idea is diagonal selection: choose a subsequence converging at the first point, refine it to converge at the second point, refine again, and then take the diagonal subsequence. The definition below names the size condition that makes this sequential construction possible. [definition: Countable Domain] A set $X$ is countable if there exists an injective function $\iota:X \to \mathbb{N}$. [/definition] Countability matters because the diagonal argument must satisfy countably many pointwise convergence requirements with one subsequence, not a different subsequence for each point. At each listed point, pointwise boundedness gives scalar subsequential compactness; the obstruction is to coordinate these infinitely many refinements without losing convergence already obtained at earlier points. The diagonal construction is the mechanism that resolves this obstruction. [quotetheorem:8531] The theorem is useful precisely because it makes no continuity assumption and no norm assumption. It is a compactness statement in the [product topology](/page/Product%20Topology) on scalar values over a countable index set. Its limitation is that countability is doing real work. [example: Diagonal Selection on the Rational Points] Let $(f_k)_{k=1}^{\infty}$ be pointwise bounded, with each $f_k:\mathbb{Q}\cap[0,1]\to\mathbb{R}$. Since $\mathbb{Q}\cap[0,1]$ is countable, choose an enumeration \begin{align*} \mathbb{Q}\cap[0,1]=\{q_1,q_2,q_3,\ldots\}. \end{align*} For each fixed $i\in\mathbb{N}$, pointwise boundedness says that there is a finite constant $M_{q_i}$ such that \begin{align*} |f_k(q_i)|\le M_{q_i} \end{align*} for every $k\in\mathbb{N}$. Thus the scalar sequence $(f_k(q_i))_{k=1}^{\infty}$ is bounded for every listed rational point $q_i$. By *Diagonal Subsequence for Pointwise Bounded Sequences*, applied to the countable domain $\mathbb{Q}\cap[0,1]$, there is a strictly increasing sequence of indices \begin{align*} k_1<k_2<k_3<\cdots \end{align*} such that the subsequence $(f_{k_j})_{j=1}^{\infty}$ converges pointwise on $\mathbb{Q}\cap[0,1]$. Written out at a single rational point, this means that for every $i\in\mathbb{N}$ there is a real number $L_i$ such that \begin{align*} \lim_{j\to\infty} f_{k_j}(q_i)=L_i. \end{align*} The conclusion is exactly about the rational points in the domain: it gives convergence of $(f_{k_j}(q_i))_{j=1}^{\infty}$ for each $q_i\in\mathbb{Q}\cap[0,1]$, and it makes no assertion at irrational points because the functions are not defined there. [/example] The rational example is a common first step in compactness proofs on intervals. If additional continuity information is available, convergence on a dense countable subset may propagate to convergence on the whole interval. Pointwise boundedness supplies the scalar compactness; equicontinuity supplies the propagation. ## Equicontinuity and Compactness Pointwise boundedness alone allows moving spikes and discontinuous limits. To prevent this, we need a condition saying that the functions do not oscillate at smaller and smaller scales in a way that depends on the index. Equicontinuity is the standard such condition. [definition: Equicontinuous Sequence of Functions] Let $(X,d_X)$ be a metric space, let $(Y,d_Y)$ be a metric space, and let $(f_k)_{k=1}^{\infty}$ be a sequence of functions $f_k:X \to Y$. The sequence $(f_k)_{k=1}^{\infty}$ is equicontinuous on $X$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in X$ and all $k\in\mathbb{N}$, \begin{align*} d_X(x,y)<\delta &\implies d_Y(f_k(x),f_k(y))<\varepsilon. \end{align*} [/definition] Equicontinuity turns a pointwise bound at one nearby point into bounds in a neighbourhood, uniformly over the sequence. On a compact domain, finitely many neighbourhoods cover the whole space. The next theorem makes this upgrade precise and explains why pointwise boundedness is often paired with equicontinuity in compactness arguments. [quotetheorem:8534] This theorem explains why the moving spike example fails to be equicontinuous. The spike heights grow while their widths shrink, so no single $\delta$ controls the oscillation for all $k$. [example: Lipschitz Control Gives Equicontinuity] Let $K=[0,1]$, and suppose that there is a constant $0\le L<\infty$ such that, for every $x,y\in K$ and every $k\in\mathbb{N}$, \begin{align*} |f_k(x)-f_k(y)|\le L|x-y|. \end{align*} First this common Lipschitz bound gives equicontinuity. If $L=0$, then $|f_k(x)-f_k(y)|\le 0$ for all $x,y,k$, so any $\delta>0$ works. If $L>0$ and $\varepsilon>0$, choose $\delta=\varepsilon/L$. Then $|x-y|<\delta$ implies \begin{align*} |f_k(x)-f_k(y)|\le L|x-y|<L\delta=\varepsilon \end{align*} for every $k\in\mathbb{N}$. Now assume the sequence is pointwise bounded at one point $x_0\in K$, so there is a finite constant $M_{x_0}$ with \begin{align*} |f_k(x_0)|\le M_{x_0} \end{align*} for every $k\in\mathbb{N}$. For any $x\in K$ and any $k\in\mathbb{N}$, the triangle inequality gives \begin{align*} |f_k(x)|=|f_k(x)-f_k(x_0)+f_k(x_0)|\le |f_k(x)-f_k(x_0)|+|f_k(x_0)|. \end{align*} Using the Lipschitz estimate with $y=x_0$, \begin{align*} |f_k(x)-f_k(x_0)|\le L|x-x_0|. \end{align*} Since $x,x_0\in[0,1]$, we have $|x-x_0|\le 1$, and therefore \begin{align*} |f_k(x)|\le L|x-x_0|+M_{x_0}\le L+M_{x_0}. \end{align*} Thus one bound at a single point, together with a Lipschitz constant independent of $k$, gives a single uniform bound on all values $f_k(x)$ over $x\in[0,1]$ and $k\in\mathbb{N}$. [/example] The Lipschitz example gives an elementary version of the compact theorem. The next [compactness theorem](/theorems/2748) goes further: instead of merely producing a uniform bound, it produces a uniformly convergent subsequence when continuity and equicontinuity are present. [quotetheorem:66] Arzela-Ascoli shows that pointwise boundedness is not a weak condition by accident; it is weak because it is designed to be combined with a regularity condition. The two hypotheses control different directions: pointwise boundedness controls vertical escape, and equicontinuity controls horizontal oscillation. ## Norms, Supremum Bounds, and Function Spaces In a normed function space, boundedness of a sequence usually means boundedness in the norm. This is a different condition from pointwise boundedness unless the norm controls point evaluations in a uniform way. The next definition fixes the ambient meaning of boundedness for an abstract normed space before comparing it with pointwise boundedness. [definition: Bounded Sequence in a Normed Space] Let $(V,\|\cdot\|_V)$ be a normed space. A sequence $(v_k)_{k=1}^{\infty}$ in $V$ is bounded in $V$ if there exists $M<\infty$ such that \begin{align*} \|v_k\|_V &\le M \end{align*} for every $k\in\mathbb{N}$. [/definition] To compare pointwise boundedness with norm boundedness in spaces of bounded functions, we need a norm that measures the largest value of a function over its entire domain. This norm converts the informal phrase "one bound for all points" into a single number attached to the function. [definition: Supremum Norm] Let $X$ be a set and let $f:X\to\mathbb{R}$ or $f:X\to\mathbb{C}$ be a bounded function. The supremum norm of $f$ on $X$ is \begin{align*} \|f\|_{\infty} &= \sup_{x\in X}|f(x)|. \end{align*} [/definition] The supremum norm controls every point evaluation at once, so a bound on the numbers $\|f_k\|_{\infty}$ prevents any value f_k(x) from escaping at any input x. There is no diagonal or compactness issue in this direction: the single norm bound already contains all pointwise inequalities simultaneously. The useful formal question is whether this simple observation can be recorded as a general implication from norm boundedness to pointwise boundedness. The quoted result isolates exactly that direction, giving a reusable criterion whenever a family of bounded functions is controlled in supremum norm. [quotetheorem:8536] The converse needs extra hypotheses. In $C(K)$, equicontinuity and compactness are one route. In functional analysis, linearity and completeness provide another route through the uniform boundedness principle. ## Linear Operators and the Uniform Boundedness Principle Pointwise boundedness becomes a major theorem in functional analysis when the functions being evaluated are bounded linear operators. Here the points are vectors in a Banach space, and the sequence values lie in another normed space. The pointwise condition says that for each fixed vector, its orbit under the operators remains bounded. [definition: Pointwise Bounded Sequence of Operators] Let $X$ be a normed space, let $Y$ be a normed space, and let $(T_k)_{k=1}^{\infty}$ be a sequence of bounded linear operators $T_k\in\mathcal{L}(X,Y)$. The sequence $(T_k)_{k=1}^{\infty}$ is pointwise bounded on $X$ if for every $x\in X$ there exists $M_x<\infty$ such that \begin{align*} \|T_kx\|_Y &\le M_x \end{align*} for every $k\in\mathbb{N}$. [/definition] This definition looks formally identical to pointwise boundedness of functions, but for arbitrary functions the pointwise constants M_x may grow with x in an uncontrolled way. For bounded linear operators on a complete domain, that uncontrolled growth cannot be hidden on separate vectors independently: linearity spreads large operator norm into visible growth on nearby vectors, and completeness supplies the Baire-category compactness needed to make this visible uniformly. [quotetheorem:716] The theorem is striking because it reverses the usual implication in a setting with enough structure. For arbitrary functions, pointwise boundedness does not imply a uniform bound. For bounded linear operators on a Banach space, it does. [example: Why Completeness Matters] Let $X=c_{00}$ be the normed space of finitely supported scalar sequences with the sup norm $\|x\|_{\infty}=\sup_{j\in\mathbb{N}}|x_j|$, and define $T_k:X\to\mathbb{R}$ by $T_k(x)=kx_k$. The map $T_k$ is linear because for $x,y\in c_{00}$ and scalars $a,b$, \begin{align*} T_k(ax+by)=k(ax_k+by_k)=a(kx_k)+b(ky_k)=aT_k(x)+bT_k(y). \end{align*} For each fixed $x\in c_{00}$, choose $N\in\mathbb{N}$ such that $x_j=0$ for every $j>N$. If $k>N$, then \begin{align*} T_k(x)=kx_k=k\cdot 0=0. \end{align*} For the remaining indices $1\le k\le N$, set \begin{align*} M_x=\max\{|T_1(x)|,|T_2(x)|,\ldots,|T_N(x)|,0\}. \end{align*} This is finite because it is the maximum of finitely many real numbers, and the preceding display gives $|T_k(x)|\le M_x$ for every $k\in\mathbb{N}$. Thus the sequence $(T_k)_{k=1}^{\infty}$ is pointwise bounded on $c_{00}$. Now compute the operator norm. For every $x\in c_{00}$, \begin{align*} |T_k(x)|=|kx_k|=k|x_k|\le k\|x\|_{\infty}. \end{align*} Hence $\|T_k\|_{\mathcal{L}(X,\mathbb{R})}\le k$. Let $e^{(k)}\in c_{00}$ be the sequence whose $k$th coordinate is $1$ and whose other coordinates are $0$. Then $\|e^{(k)}\|_{\infty}=1$, and \begin{align*} |T_k(e^{(k)})|=|k\cdot 1|=k. \end{align*} Therefore $\|T_k\|_{\mathcal{L}(X,\mathbb{R})}\ge k$, so \begin{align*} \|T_k\|_{\mathcal{L}(X,\mathbb{R})}=k. \end{align*} The operator norms are unbounded because for every $M<\infty$ one can choose $k>M$, giving $\|T_k\|_{\mathcal{L}(X,\mathbb{R})}=k>M$. The missing hypothesis is completeness. Indeed, define $x^{(m)}\in c_{00}$ by $x^{(m)}_j=1/j$ for $1\le j\le m$ and $x^{(m)}_j=0$ for $j>m$. If $n>m$, then \begin{align*} \|x^{(n)}-x^{(m)}\|_{\infty}=\sup_{m<j\le n}\frac{1}{j}=\frac{1}{m+1}. \end{align*} Thus $(x^{(m)})_{m=1}^{\infty}$ is Cauchy in the sup norm. Its coordinatewise sup-norm limit is $x=(1,1/2,1/3,\ldots)$, since \begin{align*} \|x-x^{(m)}\|_{\infty}=\sup_{j>m}\frac{1}{j}=\frac{1}{m+1}. \end{align*} But $x\notin c_{00}$ because none of its coordinates is zero. Therefore $c_{00}$ is not complete in the sup norm, so this example does not contradict the uniform boundedness principle. [/example] The example shows that the Banach hypothesis is not cosmetic. Completeness prevents bounded linear operators from hiding larger and larger norm growth on smaller and smaller parts of the space. ## Integration and Almost Everywhere Variants In measure theory, pointwise language often has an almost everywhere variant. This is needed because functions in $L^p$ spaces are identified up to equality outside a set of measure zero. A sequence may fail to be pointwise bounded on a null set while still being bounded at almost every point. [definition: Almost Everywhere Pointwise Bounded Sequence] Let $(E,\mathcal{E},\mu)$ be a [measure space](/page/Measure%20Space) and let $(f_k)_{k=1}^{\infty}$ be a sequence of [measurable functions](/page/Measurable%20Functions) $f_k:E\to\mathbb{R}$ or $f_k:E\to\mathbb{C}$. The sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded $\mu$-a.e. on $E$ if there exists a measurable set $N\subset E$ with $\mu(N)=0$ such that for every $x\in E\setminus N$ there exists $M_x<\infty$ satisfying \begin{align*} |f_k(x)| &\le M_x \end{align*} for every $k\in\mathbb{N}$. [/definition] This version is natural in $L^p$ theory, but it is still not the same as having an integrable majorant. Dominated convergence requires a single function $g$ that bounds all $|f_k|$ almost everywhere and is integrable, so we isolate that stronger condition next. [definition: Dominated Sequence] Let $(E,\mathcal{E},\mu)$ be a measure space and let $(f_k)_{k=1}^{\infty}$ be a sequence of measurable functions $f_k:E\to\mathbb{R}$ or $f_k:E\to\mathbb{C}$. The sequence $(f_k)_{k=1}^{\infty}$ is dominated by a function $g:E\to[0,\infty]$ if $g$ is measurable, $g\in L^1(E,\mathcal{E},\mu)$, and \begin{align*} |f_k(x)| &\le g(x) \end{align*} for every $k\in\mathbb{N}$ and $\mu$-a.e. $x\in E$, with the exceptional null set independent of $k$. [/definition] Dominated sequences are pointwise bounded almost everywhere, but the converse can fail because the pointwise bound $M_x$ may be non-integrable or may fail to be measurable when chosen without care. The next example shows the failure with an explicit monotone family. [example: Pointwise Bounded Almost Everywhere Without an Integrable Dominator] On $E=(0,1)$ with [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^1$, define $f_k:E\to\mathbb{R}$ by \begin{align*} f_k(x)=\min\{1/x,k\}. \end{align*} For every $x\in(0,1)$ and every $k\in\mathbb{N}$, both $1/x$ and $k$ are positive, so \begin{align*} 0\le f_k(x)=\min\{1/x,k\}\le 1/x. \end{align*} Since $1/x<\infty$ for each fixed $x\in(0,1)$, the sequence $(f_k)_{k=1}^{\infty}$ is pointwise bounded everywhere, with the valid point-dependent bound $M_x=1/x$. We show that no integrable dominator exists. Suppose, for contradiction, that there were a [measurable function](/page/Measurable%20Function) $g\in L^1((0,1))$ with $g\ge 0$ and a null set $N\subset(0,1)$ such that \begin{align*} f_k(x)\le g(x) \end{align*} for every $k\in\mathbb{N}$ and every $x\in(0,1)\setminus N$. Fix $x\in(0,1)\setminus N$. Choose $k\in\mathbb{N}$ with $k>1/x$. Then $\min\{1/x,k\}=1/x$, so \begin{align*} 1/x=f_k(x)\le g(x). \end{align*} Thus \begin{align*} g(x)\ge 1/x \end{align*} for $\mathcal{L}^1$-a.e. $x\in(0,1)$. For each integer $m\ge 2$, monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_0^1 g\,d\mathcal{L}^1\ge \int_{1/m}^1 g\,d\mathcal{L}^1\ge \int_{1/m}^1 \frac{1}{x}\,d\mathcal{L}^1(x). \end{align*} Using the antiderivative $\log x$ on $(0,\infty)$, \begin{align*} \int_{1/m}^1 \frac{1}{x}\,d\mathcal{L}^1(x)=\log 1-\log(1/m)=0-(-\log m)=\log m. \end{align*} Since $\log m\to\infty$ as $m\to\infty$, the finite number $\int_0^1 g\,d\mathcal{L}^1$ would have to be at least $\log m$ for every $m\ge 2$, which is impossible. Therefore the pointwise bounded sequence $(f_k)$ has no integrable dominator. [/example] The example identifies a common mistake: pointwise boundedness is not dominated convergence in disguise. It gives finite vertical bounds, not an integrable envelope. ## Failure Modes and Common Misreadings The same phrase appears in elementary analysis, topology, measure theory, and functional analysis, but the surrounding structure determines what can be concluded. The safest approach is to translate the phrase into quantifiers before using it. [remark: Quantifier Form of Pointwise Boundedness] For a sequence $(f_k)_{k=1}^{\infty}$ of scalar-valued functions on a set $X$, pointwise boundedness on $X$ is the assertion \begin{align*} \forall x\in X\,\exists M_x<\infty\,\forall k\in\mathbb{N}: |f_k(x)|\le M_x. \end{align*} Uniform boundedness on $X$ is the assertion \begin{align*} \exists M<\infty\,\forall x\in X\,\forall k\in\mathbb{N}: |f_k(x)|\le M. \end{align*} [/remark] The quantifier order is the difference between having a bound after choosing the point and choosing a bound before seeing the point. Many examples in this chapter exploit exactly that reversal. [remark: Pointwise Bounded Does Not Mean Bounded Pointwise Limit] If $(f_k)$ is pointwise bounded and converges pointwise to $f$, then $f(x)$ is finite at each $x$. This does not imply that $f$ is a bounded function on $X$. For example, $f_k(x)=x$ on $\mathbb{R}$ is pointwise bounded as a constant sequence in $k$, and its pointwise limit $f(x)=x$ is unbounded on $\mathbb{R}$. [/remark] This remark reinforces the local nature of the word pointwise. The definition controls the index variable $k$ after $x$ is fixed; it does not control the spatial variable $x$. [remark: Bounded in $L^p$ Need Not Give Pointwise Boundedness] Norm boundedness in $L^p$ does not generally imply pointwise boundedness of chosen representatives. A sequence may concentrate mass on shrinking sets while keeping its $L^p$ norm controlled, and the pointwise behaviour can depend on representatives chosen on null sets. [/remark] This warning is especially important because $L^p$ functions are equivalence classes. Statements about individual values require representatives or an almost everywhere formulation. ## Beyond and Connected Topics Pointwise bounded sequences sit at the entrance to several larger theories. In elementary analysis, they are a stepping stone from ordinary [sequences](/page/Sequence) to sequences of functions, pointwise convergence, and [uniform convergence](/page/Uniform%20Convergence). The natural continuation is to study which properties survive under different modes of convergence. In topology and real analysis, pointwise boundedness becomes meaningful when paired with equicontinuity. Arzela-Ascoli explains how pointwise bounds plus uniform control of oscillation produce compactness in spaces of continuous functions. This is a central bridge from pointwise estimates to compactness arguments. In functional analysis, the same phrase becomes the hypothesis of the uniform boundedness principle. The difference is structural: bounded [linear operators on Banach spaces](/page/Linear%20Operators%20on%20Banach%20Spaces) cannot have unbounded norms while remaining pointwise bounded on every vector. This connects the topic to Banach spaces, Baire category, and the stability of families of linear maps. In measure theory, the almost everywhere version clarifies the gap between pointwise boundedness and domination. This distinction matters for the [dominated convergence theorem](/theorems/4), Fatou-type arguments, and convergence in $L^p$ spaces. For a course-level continuation inside Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) gives the sequence and convergence background, while [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) develops compactness and uniform convergence themes. [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions) is the natural place to see these ideas interacting with function spaces and more advanced convergence arguments. ## References 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). John B. Conway, *A Course in Functional Analysis* (1990). Gerald B. Folland, *Real Analysis* (1999).