Linear analysis is the study of infinite-dimensional vector spaces equipped with extra structure — norms, inner products, topologies — that enables analysis. In finite dimensions, linear algebra suffices: every linear map is automatically continuous, every bounded sequence has a convergent subsequence, and every subspace is closed. None of these properties survive the passage to infinite dimensions, and the central challenge of the subject is understanding precisely what does and does not carry over.

The motivation is concrete. Differential equations — ordinary and partial — are naturally formulated as operator equations $Lu = f$, where $L$ is a differential operator acting on a space of [functions](/page/Function). To apply the tools of linear algebra (existence of solutions, spectral decompositions, perturbation theory), we need the space of functions to be a vector space, and $L$ to be a linear map between such spaces. But to do *analysis* — to take [limits](/page/Limit), extract convergent subsequences, approximate rough objects by smooth ones — we need more: we need a topology compatible with the algebraic structure, and ideally completeness. The theory developed in these notes provides exactly this framework.

The notes follow the Cambridge Part IB Linear Analysis course. We begin with [normed vector spaces](/page/Normed%20Vector%20Space) and their basic theory (including dual spaces and adjoint maps), then study finite-dimensional spaces (where everything is controlled by compactness of the unit ball), and proceed to the three pillars of the subject: the Hahn-Banach theorem (which guarantees a rich supply of linear functionals), the Baire Category theorem (which yields uniform boundedness, the open mapping theorem, and the closed graph theorem), and the theory of compact operators on [Hilbert spaces](/page/Hilbert%20Space) (culminating in the spectral theorem).

# Normed Vector Spaces

The starting point of linear analysis is to combine the algebraic structure of a vector space with a notion of "size" — a norm — that allows us to measure distances, define convergence, and do analysis. This chapter introduces normed vector spaces and develops the basic constructions: [Banach spaces](/page/Banach%20Space) (complete normed spaces), bounded [linear maps](/page/Linear%20Map), dual spaces, adjoints, and the bidual. These objects form the vocabulary of the entire subject.

## The Norm and its Topology

A vector space on its own carries no analytic structure: there is no notion of convergence, no concept of [continuity](/page/Continuity), no way to say that two elements are "close." A norm remedies this by assigning a non-negative real number to each vector, measuring its size.

[definition:Normed Vector Space] A **normed vector space** is a pair $(V, \|\cdot\|)$ where $V$ is a vector space over a field $\mathbb{F}$ (either $\mathbb{R}$ or $\mathbb{C}$) and $\|\cdot\|: V \to \mathbb{R}$ is a **norm**, i.e. a function satisfying: (i) $\|v\| \ge 0$ for all $v \in V$, with equality if and only if $v = 0$. (ii) $\|\lambda v\| = |\lambda| \cdot \|v\|$ for all $\lambda \in \mathbb{F}$, $v \in V$. (iii) $\|v + w\| \le \|v\| + \|w\|$ for all $v, w \in V$ (the triangle inequality). [/definition]

A norm induces a metric $d(v, w) := \|v - w\|$, which in turn induces a topology: a set $U \subset V$ is open if and only if every point $x \in U$ has an open ball $B_\varepsilon(x) := \{y \in V : \|y - x\| < \varepsilon\} \subset U$ around it. This gives $V$ the structure of a [topological](/page/Topology) space, and a natural question arises: how does the topology interact with the algebraic operations?

The answer is that the algebraic operations are continuous with respect to the norm topology. This is the content of the following result.

[theorem:Continuity of Algebraic Operations] Let $(V, \|\cdot\|)$ be a normed vector space. Then addition $+: V \times V \to V$ and scalar multiplication $\cdot: \mathbb{F} \times V \to V$ are continuous with respect to the norm topology on $V$ (and the product topology on the domain). [/theorem]

The key point is that the norm provides *quantitative* control over both operations. For addition, this is immediate from the triangle inequality: if $w_i \in B_{\varepsilon/2}(v_i)$ for $i = 1, 2$, then $\|w_1 + w_2 - v_1 - v_2\| \le \|w_1 - v_1\| + \|w_2 - v_2\| < \varepsilon$, so the sum stays in $B_\varepsilon(v_1 + v_2)$. For scalar multiplication, the argument requires a bit more care because both the scalar and the vector can vary simultaneously, but the essential mechanism is the same: the estimate $\|aw - \lambda v\| \le |a| \cdot \|w - v\| + |a - \lambda| \cdot \|v\|$ shows that both factors can be controlled independently.

[proof] **Step 1: Continuity of addition.** Let $U \subset V$ be open and take $(v_1, v_2)$ with $v_1 + v_2 \in U$. Since $U$ is open, there exists $\varepsilon > 0$ with $B_\varepsilon(v_1 + v_2) \subset U$. For any $w_i \in B_{\varepsilon/2}(v_i)$, the triangle inequality gives: \begin{align*} \|w_1 + w_2 - v_1 - v_2\| \le \|w_1 - v_1\| + \|w_2 - v_2\| < \varepsilon/2 + \varepsilon/2 = \varepsilon. \end{align*} Hence $B_{\varepsilon/2}(v_1) \times B_{\varepsilon/2}(v_2) \subset (+)^{-1}(U)$, which is open in the product topology. **Step 2: Continuity of scalar multiplication.** Let $U \subset V$ be open and take $(\lambda, v)$ with $\lambda v \in U$. Choose $\varepsilon > 0$ with $B_\varepsilon(\lambda v) \subset U$. *Case $\lambda \ne 0$, $v \ne 0$:* Set $\delta = \varepsilon / (2\|v\|)$ and choose $r > 0$ small enough that $|a| r < \varepsilon/2$ for all $a \in (\lambda - \delta, \lambda + \delta)$. For $a \in (\lambda - \delta, \lambda + \delta)$ and $w \in B_r(v)$: \begin{align*} \|aw - \lambda v\| \le \|aw - av\| + \|av - \lambda v\| = |a| \cdot \|w - v\| + |a - \lambda| \cdot \|v\| < \varepsilon/2 + \varepsilon/2 = \varepsilon. \end{align*} *Case $\lambda = 0$:* We need $\|aw\| < \varepsilon$. For $w \in B_1(v)$ we have $\|w\| < \|v\| + 1$, so taking $|a| < \varepsilon / (\|v\| + 1)$ gives $\|aw\| = |a| \cdot \|w\| < \varepsilon$. *Case $v = 0$:* We need $\|aw\| < \varepsilon$. Taking $w \in B_r(0)$ and $a \in (\lambda - \delta, \lambda + \delta)$ with $r$ small enough that $(|\lambda| + \delta) r < \varepsilon$ gives $\|aw\| = |a| \cdot \|w\| < (|\lambda| + \delta)r < \varepsilon$. In every case we have found an open neighbourhood of $(\lambda, v)$ mapping into $U$. [/proof]

This result shows that every normed vector space is a **[topological vector space](/page/Topological%20Vector%20Space)** — a vector space equipped with a topology making addition and scalar multiplication continuous, and in which singleton sets are closed. The converse question — when is a topological vector space normable? — has a clean answer: a topological vector space admits a norm inducing its topology if and only if it has an absolutely convex bounded neighbourhood of $0$. We will not pursue this direction further, but it explains why normed spaces occupy a privileged position: they are precisely the topological vector spaces whose local geometry is controlled by a single convex, balanced set (the unit ball).

## Banach Spaces

A normed vector space provides the setting for analysis, but without completeness — the property that every Cauchy sequence converges — many fundamental tools are unavailable. We cannot guarantee that limits of approximating [sequences](/page/Sequence) exist in the space, we cannot take closures meaningfully, and we cannot apply the powerful results (Baire category, open mapping, uniform boundedness) that form the backbone of the subject. Completeness is therefore not a luxury but a necessity.

[definition:Banach Space] A normed vector space $(V, \|\cdot\|)$ is a **Banach space** if it is complete as a metric space, i.e. if every Cauchy sequence in $V$ converges to a limit in $V$. [/definition]

[example:Bounded Functions] Let $X$ be a compact Hausdorff space. Define \begin{align*} B(X) := \{f: X \to \mathbb{R} \mid f \text{ is bounded}\} \end{align*} with the supremum norm $\|f\|_\infty := \sup_{x \in X} |f(x)|$. We claim $(B(X), \|\cdot\|_\infty)$ is a Banach space. Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $B(X)$. For each fixed $x \in X$, the numerical sequence $(f_n(x))_{n=1}^\infty$ satisfies $|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty$, so it is Cauchy in $\mathbb{R}$. By completeness of $\mathbb{R}$, the pointwise limit $f(x) := \lim_{n \to \infty} f_n(x)$ exists for every $x$. We now show $f_n \to f$ in $B(X)$. Let $\varepsilon > 0$. Since $(f_n)$ is Cauchy, there exists $N$ such that $\|f_n - f_m\|_\infty < \varepsilon/2$ for all $n, m \ge N$. For each fixed $x \in X$ and $n \ge N$, take $m \to \infty$ in the inequality $|f_n(x) - f_m(x)| < \varepsilon/2$ to obtain $|f_n(x) - f(x)| \le \varepsilon/2 < \varepsilon$. Since this holds for all $x$, we have $\|f_n - f\|_\infty \le \varepsilon$ for all $n \ge N$. Finally, $f$ is bounded because $|f(x)| \le |f(x) - f_N(x)| + |f_N(x)| \le \varepsilon + \|f_N\|_\infty$ for all $x$. [/example]

The argument above — construct the limit pointwise using completeness of the scalars, then upgrade to norm convergence — is the standard method for proving completeness of function spaces defined by supremum-type norms. It applies equally to the next example.

[example:Continuous Functions on a Compact Space] Let $X$ be a compact Hausdorff space. Define $C(X) := \{f: X \to \mathbb{R} \mid f \text{ is continuous}\}$ with the supremum norm. Since $X$ is compact, every continuous $f: X \to \mathbb{R}$ is bounded, so $C(X) \subset B(X)$ is a linear subspace. The space $(C(X), \|\cdot\|_\infty)$ is a Banach space. Since $(B(X), \|\cdot\|_\infty)$ is complete, it suffices to show that $C(X)$ is a closed subspace. Suppose $f_n \to f$ in $B(X)$ with each $f_n$ continuous. Then $f_n \to f$ uniformly, and the uniform limit of continuous functions is continuous. Hence $f \in C(X)$, so $C(X)$ is closed and therefore complete. [/example]