The Fundamental Theorems

Infinite-dimensional analysis fails in ways finite-dimensional analysis does not. A linear functional on every finite-dimensional space is automatically continuous; on an infinite-dimensional normed space it need not be. A linear operator between finite-dimensional spaces is automatically bounded, and on infinite-dimensional ones it need not be. Compactness, closedness and continuity come apart in ways that have no finite-dimensional analogue.

The four theorems in this section are the repairs. Each one says that a plausible pathology cannot occur, provided the spaces are complete:

• an everywhere-defined linear functional on a Banach space is bounded;
• a surjective bounded operator between Banach spaces is open;
• an everywhere-defined linear operator between Banach spaces with a closed graph is bounded;
• a pointwise bounded family of bounded operators is uniformly bounded.

Completeness is not decoration. Every one of these is false on an incomplete space, and section 4.9 shows why: the proofs all consume the same ingredient, Baire’s theorem, which is a theorem about complete metric spaces and nothing else.

The shared idea is this. Completeness means a convergent Cauchy sequence lands somewhere in the space. Baire’s theorem turns that single property into: a complete space cannot be a countable union of thin sets. Once you have that, the three Baire-based theorems below all follow from the same three moves — assume a pathology, write the space as a countable union of closed sets, invoke Baire to find one with interior, and use the interior to bootstrap a uniform estimate out of a pointwise one.

This is the standard “principle of uniform boundedness” pattern, and seeing it three times in a row is the point of the section. What makes it feel like a trick the first time and like a tool the third time is that the pointwise-to-uniform gap is exactly what completeness closes.

Definition. A subset $A$ of a topological space is nowhere dense if its closure has empty interior. A space is meagre (or of first category) if it is a countable union of nowhere dense sets.

Theorem 4.1 (Baire Category Theorem). A complete metric space is not meagre in itself. Equivalently, if $\{F_n\}_{n \geq 1}$ are closed subsets of a complete metric space and their interiors are dense, then $\bigcap_n F_n$ is dense — and in particular nonempty.

Proof. Let $(x_n)$ be a Cauchy sequence and suppose all $F_n$ are closed with dense interior. Define

Each $N_n$ is closed, and $N_n$ has empty interior: $N_{n-1}^\circ$ is dense and disjoint from $N_n$, because a point of $N_n$ is within $2^{-n}$ of some $x_{n-1}$, so a small ball about any point of $N_{n-1}^\circ$ avoids $N_n$. Hence $N_n$ is nowhere dense.

Since $X$ is complete and not meagre, $X \neq \bigcup_n N_n$, so there is $y \in X \setminus \bigcup_n N_n$. As $y \notin N_1$ there is $x_1 \in F_1$ with $d(y, x_1) < 1/2$. Having $x_1, \dots, x_{n-1}$ with $d(x_{n-1}, y) < 2^{-(n-1)}$ and $y \notin N_n$, there is $x_n \in F_n$ with $d(x_n, y) < 2^{-n}$. So $(x_n)$ is Cauchy and lies in $F_n$ for all $n$.

For each $n$, $F_n$ is closed and contains all $x_m$ for $m \geq n$, so $F_n \ni \lim_{m\to\infty} x_m = y$. Completeness is used precisely here, and nowhere else. Hence $y \in \bigcap_n F_n$. $\blacksquare$

That completeness is load-bearing is not a rhetorical point: an incomplete metric space can be meagre in itself. $\mathbb{Q} \subset \mathbb{R}$ is a countable union of singletons, each nowhere dense. So Baire fails, and with it the open mapping theorem. This is why “Banach” appears in every hypothesis below.

Theorem 4.2 (Hahn-Banach, analytic form). Let $X$ be a real vector space and $p : X \to \mathbb{R}$ sublinear, meaning $p(x+y) \leq p(x) + p(y)$ and $p(tx) = tp(x)$ for $t \geq 0$. Let $M \leq X$ be a subspace and $f : M \to \mathbb{R}$ linear with $f(m) \leq p(m)$ for all $m \in M$. Then there is a linear $F : X \to \mathbb{R}$ with $F|_M = f$ and $F(x) \leq p(x)$ for all $x \in X$.

Proof (one-dimensional extension, then Zorn). Given $x_0 \notin M$, we seek $\alpha \in \mathbb{R}$ so that $F(m + \lambda x_0) = f(m) + \lambda\alpha$ stays below $p$ on all of $M + \mathbb{R}x_0$. The constraint $f(m) + \lambda\alpha \leq p(m + \lambda x_0)$ must hold for every $m \in M$ and every $\lambda \in \mathbb{R}$, so

the inequality flipping when dividing by a negative number. Hence we need $\alpha \in [a, b]$ where

Sublinearity gives $a \leq b$. To see it, fix $\lambda > 0$ and write $\lambda = \lambda' + \lambda''$ with $\lambda', \lambda'' > 0$. By sublinearity of $p$ and linearity of $f$,

Dividing by $\lambda'' > 0$, the $\lambda''$-quotient is at most $b$ by definition, and the $\lambda'$-quotient is at least $a$. Since $a \leq \frac{p(m+\lambda x_0) - f(m)}{\lambda} - \frac{f(m+\lambda'x_0)-f(m)}{\lambda''}$ and the right-hand term is at most $b$, we get $a \leq b$. Choose $\alpha \in [a,b]$; then $F$ on $M + \mathbb{R}x_0$ is linear and bounded above by $p$.

Order the set of pairs $(N, F)$ where $N$ is a subspace with $M \leq N \leq X$ and $F$ is linear on $N$ with $F|_M = f$ and $F \leq p$ on $N$, ordered by extension. This set is inductive by the one-dimensional step. By Zorn there is a maximal element $(N, F)$. If $N \neq X$, pick $x_0 \in X \setminus N$; the one-dimensional step produces a strictly larger element, a contradiction. So $N = X$. $\blacksquare$

Theorem 4.3 (Hahn-Banach, normed form). If $M \leq X$ is a subspace of a normed space and $f \in M^*$, then $f$ extends to $F \in X^*$ with $\|F\| = \|f\|$.

Proof. Apply 4.2 with $p(x) = \|f\|\,\|x\|$. Then $f(m) \leq |f(m)| \leq \|f\|\,\|m\| = p(m)$, so the hypothesis holds, and we get $F \leq p$. Applying the same to $-f$ gives $-F \leq p$, so $|F(x)| \leq p(x) = \|f\|\,\|x\|$, i.e. $\|F\| \leq \|f\|$. Since $F$ extends $f$, $\|f\| \leq \|F\|$, hence $\|F\| = \|f\|$. $\blacksquare$

Corollary 4.4 (separating functionals). For $x_0 \neq 0$ there is $\varphi \in X^*$ with $\|\varphi\| = 1$ and $\varphi(x_0) = \|x_0\|$.

Proof. Let $M = \mathbb{R}x_0$ and $f(\lambda x_0) = \lambda\|x_0\|$. Then $f \leq \|\cdot\|$ on $M$ and $\|f\| = 1$; extend. $\blacksquare$

Consequence. $X^*$ separates points, so $\{0\}$ is the only vector annihilated by every $\varphi \in X^*$. Equivalently, the canonical embedding $J : X \to X^{**}$, $Jx(\varphi) = \varphi(x)$, is injective. Recall from Bounded Linear Operators that $J$ is an isometry, so it is also a homeomorphism onto its image. For a Hilbert space $J$ is onto — that is the Riesz representation theorem. For a general Banach space it is onto only for reflexive spaces, and the Hahn-Banach corollary is exactly the “one half” that always holds.

Hahn-Banach builds the objects the theory needs. On $\ell^\infty$ consider

where $\lim_{n\to\sup}$ means $\limsup$. This is bounded: $|s_n| \leq n\|x\|_\infty$, and $|L| \leq \limsup |s_n|$ is finite but not $\leq \|x\|_\infty$, so $L$ does not yet dominate $\|\cdot\|_\infty$ and Hahn-Banach cannot be applied directly to it.

The fix is to define the limiting set $A = \{y : y_1 \geq 0,\ y_{n+1} \geq y_n - 1, \ \liminf (y_1 + \dots + y_n) \geq 0\}$ in a larger space, show $A$ is convex and does not absorb $n$, and apply Hahn-Banach to obtain $\varphi$ dominated by $\|\cdot\|_\infty$ and dominating $L$. Then $\mathrm{LIM}(x) = \varphi(L(x))$ extends the ordinary limit, is linear, satisfies $\mathrm{LIM}(x) = \mathrm{LIM}(Sx)$ for the shift $S$, and is positive. Such an object cannot be a multiple of the usual limit: it agrees with the limit on convergent sequences yet is non-zero on others, which is possible precisely because $\ell^\infty$ has no continuous extension of the limit functional. That non-uniqueness is the phenomenon Banach limits were invented to exhibit.

Theorem 4.5 (Open Mapping, surjective form). Let $X, Y$ be Banach spaces and $T \in \mathcal{B}(X, Y)$ surjective. Then there is $C > 0$ with

Theorem 4.6 (Open Mapping). Under the same hypotheses $T$ maps open sets to open sets.

Proof. Set $Y_n = \overline{T(B_X(0, 1/n))}$ for $n \geq 1$. Surjectivity gives $Y = T(X) \subseteq \bigcup_n Y_n$.

Each $Y_n$ is closed: if $y_k \to y$ with $y_k \in Y_n$ and $y_k = \lim_j T(x_{k,j})$ for $x_{k,j} \in B_X(0, 1/n)$, then $(x_{k,j})_{j}$ is Cauchy, so $X$ being complete gives $x_{k,j} \to x_k$ with $\|x_k\| \leq 1/n$ and $T(x_k) = y_k$. Hence $x_k \to x$ with $x \in \overline{B_X(0,1/n)}$ and $Tx = y$, so $y \in Y_n$.

By Baire, some $Y_n$ has nonempty interior: $Y_m - Y_m \subseteq Y_{2m}$, so $Y_n - Y_n$ contains $Y_n - y_0$ for $y_0 \in Y_n^\circ$, and $Y_n - y_0$ contains a ball $B_Y(y_0, \delta)$ with $y_0 \in Y_n$. Thus $B_Y(0, \delta) \subseteq Y_n - Y_n \subseteq Y_{2n}$.

Write $u = a - b$ with $a, b \in Y_{2n}$, so $\|a\|, \|b\| \leq 1/(2n)$ and $a - b = Tx$ for some $x \in B_X(0, 1/n)$. Given $y$ with $\|y\| < \delta/2^{k}$ for some $k$, choose $a_1, b_1 \in Y_{4n}$ with $y = a_1 - b_1$ and $\|a_1\|, \|b_1\| < 1/(4n)$, so $y = Tx_1$ with $\|x_1\| < 1/(2n)$. Then $y/2 = (a_1 - b_1)/2$, and

with $a_1 - a_1/2 = a_1/2 \in Y_{4n} \subseteq Y_{2n}$ and $a_1/2 - b_1 = (a_1 - 2b_1)/2 \in Y_{2n}$ because $a_1 \in Y_{4n} \subseteq Y_{2n}$ and $2b_1 \in Y_{2n}$, so $a_1 - 2b_1 \in Y_{2n}$. Thus $y/2 = Tx'$ with $\|x'\| < 1/n$. Applying this $k$ times, $y = Tx$ with $\|x\| < 2^k/n$.

Since $\|y\| < \delta/2^k$ and $\|x\| < 2^k/n$, we get $\|x\| < 2\delta^{-1}\|y\|$. Taking $C = 2/\delta$ gives $B_Y(0,1) \subseteq T(B_X(0, C))$. This is 4.5.

For 4.6, let $U \subseteq X$ be open and $u_0 \in U$. With $r = d(u_0, X \setminus U) > 0$, $u_0 + B_X(0, r) \subseteq U$, so $T(u_0) + T(B_X(0, r)) \subseteq T(U)$. For $C$ as above, $B_Y(0, r/C) \subseteq T(B_X(0, r))$, so $T(U)$ contains a ball about $T(u_0)$, i.e. $T(U)$ is open. $\blacksquare$

Corollary 4.7 (Bounded Inverse). If $T \in \mathcal{B}(X, Y)$ is bijective and $X, Y$ are Banach, then $T^{-1} \in \mathcal{B}(Y, X)$.

Proof. By 4.6, $T$ is open. So $T^{-1}$ is continuous at $0$, hence everywhere; and linear. Its norm is $\|T^{-1}\| = \sup_{\|y\| \leq 1}\|T^{-1}y\| \leq C < \infty$. $\blacksquare$

Theorem 4.8. Let $X$ be a vector space that is Banach under both norms $\|\cdot\|_1$ and $\|\cdot\|_2$. Suppose for some $C > 0$, $\|x\|_1 \leq C\|x\|_2$ for all $x \in X$. Then $\|x\|_2 \leq C'\|x\|_1$ for some $C' > 0$; the norms are equivalent.

Proof. Let $I : (X, \|\cdot\|_2) \to (X, \|\cdot\|_1)$ be the identity. It is linear and bijective, and $\|Ix\|_1 \leq C\|Ix\|_2$, so $I \in \mathcal{B}(X_2, X_1)$. Both spaces are Banach, so 4.7 gives $I^{-1} \in \mathcal{B}(X_1, X_2)$ with $\|I^{-1}\| = C' < \infty$. Then $\|x\|_2 \leq C'\|x\|_1$. $\blacksquare$

The completeness hypothesis on both norms is essential, and it is easy to misread the theorem as saying that one domination suffices. It does not. If $X$ is complete under only $\|\cdot\|_1$, one inequality gives a continuous bijection whose inverse may be discontinuous — the canonical embedding of $c_0$ into $\ell^\infty$, or the identity from $\ell^1$ to $c_0$, are the standard examples. Equivalence requires both directions, and it is completeness that produces the second one.

Theorem 4.9 (Closed Graph). Let $X, Y$ be Banach spaces and $T : X \to Y$ linear (no boundedness assumed). Then $T$ is bounded if and only if its graph $\Gamma(T) = \{(x, Tx)\}$ is closed in $X \times Y$.

Proof ($\Rightarrow$). A continuous map has closed graph when the codomain is Hausdorff: if $(x_n, Tx_n) \to (x, y)$ then $Tx_n \to Tx$ by continuity and also $\to y$, so $y = Tx$.

($\Leftarrow$) Suppose $\Gamma(T)$ is closed. The naive route — restricting $T$ to $\{x : Tx \in \overline{B_Y(0,1)}\}$ and recursing — is circular, because the recursive instance is a closed graph theorem in its own right. The way out is to make the restricted domain complete.

Let $X_0 = \{x \in X : Tx \in \overline{B_Y(0,1)}\}$, a subspace, and let $\overline{X_0}$ be its closure in $X$ (so $\overline{X_0}$ is Banach). Define $\hat{T} : \overline{X_0} \to Y$ on the closure by $x_k \to x \in \overline{X_0}$ implies $\hat{T}x = \lim Tx_k$. This limit exists: $Tx_k$ is Cauchy in $Y$, and $Y$ is complete; and it is independent of the approximating sequence, since if $x_k \to x$ and $y_k \to x$ then $(x_k - y_k) \to 0$, so $T(x_k - y_k) \to 0$ by continuity of $T$, and $\lim Tx_k = \lim Ty_k$. So $\hat{T}$ is well defined and continuous, and it extends $T$ on $X_0$. It is surjective, because $T$ is surjective and every $y \in Y$ has $Tx = y$ for some $x$, hence $y \in T(\overline{X_0})$.

Both $\overline{X_0}$ and $Y$ are Banach, so 4.5 gives $C$ with $B_Y(0,1) \subseteq \hat{T}(B_{\overline{X_0}}(0, C))$. Since $B_X(0, C) \subseteq B_{\overline{X_0}}(0, C)$, we get $B_Y(0,1) \subseteq T(B_X(0, C))$ and therefore $\|Tx\| \leq C\|x\|$ for $x \in X_0$.

Finally, $X = \bigcup_{n\geq 1} nX_0$: for any $x \in X$ take $n > 1/\|x\|$, so $x/n \in B_X(0,1) \subseteq X_0$ and $x \in nX_0$. A vector space cannot be a finite union of proper subspaces — pick the largest index $n$ with $nX_0 \neq X$ and a point outside it, and it lies in none of the others — so $X_0 = X$. Hence $\|Tx\| \leq C\|x\|$ on all of $X$. $\blacksquare$

Theorem 4.10 (Closed Range). For $T \in \mathcal{B}(X, Y)$ with $X, Y$ Banach, these are equivalent: (i) $R(T)$ is closed in $Y$; (ii) $R(T^*)$ is closed in $X^*$; (iii) there is $c>0$ such that every $y \in R(T)$ has $x$ with $Tx = y$ and $\|x\| \leq c\|y\|$.

Proof sketch. (iii) $\Rightarrow$ (i) is immediate from the estimate. For (i) $\Rightarrow$ (iii), $T$ is open onto $R(T)$, giving (iii) with the norm of $R(T)$; for (ii) $\Rightarrow$ (i), apply the open mapping theorem to $T^*$ between $X^*$ and $R(T^*)^\perp$ and use $\overline{R(T^*)} = (T(X))^\perp$. $\blacksquare$

Theorem 4.11 (Banach-Steinhaus). Let $X$ be a Banach space, $Y$ a normed space, and $\{T_\alpha\}_{\alpha \in A} \subseteq \mathcal{B}(X, Y)$ with $\sup_\alpha \|T_\alpha x\| < \infty$ for each $x \in X$. Then $\sup_\alpha \|T_\alpha\| < \infty$.

Proof. Let $E_n = \{x \in X : \sup_\alpha \|T_\alpha x\| \leq n\}$. By hypothesis $X = \bigcup_{n\geq1} E_n$. Each $E_n$ is closed: if $x_k \to x$ with $x_k \in E_n$ then $T_\alpha x_k \to T_\alpha x$ for each $\alpha$, so $\|T_\alpha x\| \leq n$.

By Baire, some $E_n$ has nonempty interior, so $B_X(x_0, r) \subseteq E_n$ for some $x_0$ and $r > 0$. Given $x \in B_X(0, r)$, both $x_0 + x$ and $x_0$ lie in $E_n$, so by linearity

for every $\alpha$. Rescaling, $\|T_\alpha\| \leq \frac{2n}{r}$. $\blacksquare$

This is the Baire pattern in its purest form, and it is worth being explicit about where each hypothesis is used. Pointwise boundedness gives the cover $X = \bigcup_n E_n$; linearity makes each $E_n$ closed; completeness gives Baire; Baire gives an interior point; and the interior is what turns a bound that may depend on $x$ into one that does not.

Corollary 4.12 (Resonance Theorem). If $\{T_n\} \subseteq \mathcal{B}(X, Y)$ and $\lim_n T_n x$ exists for every $x \in X$, then $T x = \lim_n T_n x$ defines $T \in \mathcal{B}(X, Y)$.

Proof. A convergent sequence is bounded, so $\sup_n \|T_n x\| < \infty$ for each $x$, and 4.11 applies to give $\sup_n \|T_n\| = C$. Then $T$ is linear (interchange limits and finite sums) and

so $\|Tx\| \leq C\|x\|$, i.e. $T \in \mathcal{B}(X,Y)$. $\blacksquare$

Fourier series diverge for some continuous functions. Let $S_n f$ be the $n$-th partial sum operator on $C[-\pi,\pi]$. If every continuous $f$ had $\|S_n f\|_\infty$ bounded for each fixed $f$, 4.11 would give $\sup_n \|S_n\| < \infty$. But $S_n f = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) D_n(t)\,dt$ with $D_n(t) = \sum_{k=-n}^n e^{ikt}$, so

So $\sup_n \|S_n\| = \infty$, and by contraposition some $f \in C[-\pi,\pi]$ has $\sup_n \|S_n f\|_\infty = \infty$: a continuous function whose Fourier series diverges. This is the classical existence proof, and it is worth noticing what it needed — no explicit bad function, only the failure of a uniform bound that the theorem makes unavoidable.

Weakly bounded implies norm bounded. For a single family, $\sup_\alpha |\langle T_\alpha x, y\rangle| < \infty$ for all $x, y$ implies $\sup_\alpha \|T_\alpha\| < \infty$. Weak convergence is weaker than norm convergence at the level of sequences, but for families the two notions of boundedness coincide on a Banach domain. That contrast is the content of Weak and Weak* Convergence.

Higher derivatives are unbounded, and nowhere-differentiable functions are generic. Let $D^n : C^1[0,1] \to C[0,1]$, $D^n f = f^{(n)}$. With the norm $\|f\|_\infty$ on $C^1[0,1]$, each $D^n$ is unbounded: take $f(x) = x^n / n!$, so $\|f\|_\infty = 1/n!$ while $\|D^n f\|_\infty = 1$. The operator norms blow up like $n!$. This is a fact about the domain, not the operator — on $C^1$ with $\|f\|_\infty + \|f'\|_\infty$, $D$ is bounded with norm $1$. Which norm you equip a space with is part of specifying an operator.

The same Baire argument that proves uniform boundedness produces the classical consequence. Let $E_n$ be the set of continuous $f$ that are not differentiable on any interval of length $> 1/n$. Each $E_n$ is closed, and each is dense: a polynomial plus a small multiple of the Weierstrass nowhere-differentiable function stays inside $E_n$ while being arbitrarily close to the polynomial. Baire gives $\bigcap_n E_n$ dense, so nowhere-differentiable continuous functions are residual — generic, not exotic.

They are not four independent facts.

• Hahn-Banach is the odd one out: it needs no Baire and no completeness. Its content is that $X^*$ is large enough to separate points.
• Open mapping, closed graph and uniform boundedness are all Baire, and all three share the skeleton: write the space as a countable union of closed sets indexed by a bound, let Baire select a set with interior, and use the interior to upgrade a pointwise estimate to a uniform one.
• Closed graph is derived from open mapping in 4.4, by completing the domain and then showing the domain was already everything.
• Hahn-Banach + the open mapping theorem give the closed range theorem, and Hahn-Banach + reflexivity give the Riesz representation theorem.

The one-line summary: completeness plus Baire is what stops a pointwise statement from being a uniform one, and the three Baire-based theorems are three faces of that single fact.

Weak solutions of the Dirichlet problem. Let $\Omega$ be bounded. The operator $A : H^1_0(\Omega) \to H^{-1}(\Omega)$, $Au = -\Delta u$ with zero trace, is bounded, and the form $a(u, v) = \int_\Omega \nabla u \cdot \nabla v$ is continuous and coercive on $H^1_0(\Omega)$:

with $\lambda > 0$ by Poincare. The Lax-Milgram lemma — which is Hahn-Banach for coercivity plus the Hilbert-space geometry — gives a unique $u \in H^1_0(\Omega)$ with $a(u,v) = \langle f, v\rangle$ for every $v$. Existence, uniqueness and continuous dependence on $f$ all come from the functional-analytic theorems, not from regularity estimates on $\Delta$.

Neumann series. If $T \in \mathcal{B}(X)$ with $\|T\| < 1$, then $I - T$ is invertible with

because the series converges in operator norm (so $\|S - T^n S\| \to 0$ by comparing geometric series) and $I - T$ is injective: $(I-T)x = 0$ gives $x = Tx$, so $\|x\| \leq \|T\|\|x\| < \|x\|$ unless $x = 0$; surjectivity then follows from 4.7. This is the resolvent of a bounded operator off the closed unit disc, and it is what makes spectral theory for bounded operators work.

Compactness of the embedding. In finite dimensions every bounded set is compact. The failure of that in infinite dimensions — the unit ball of $\ell^2$ has no convergent subsequence — is what makes compactness arguments, and hence the Arzela-Ascoli and Rellich-Kondrachov theorems, carry real content here. The repairs are exactly the theorems above: a weakly convergent sequence that is bounded has a weakly convergent subsequence in a separable reflexive space, and the boundedness comes from uniform boundedness applied to the evaluation functionals.

Problem 1. Let $M$ be a closed subspace of a Hilbert space $H$. Show that $H = M \oplus M^\perp$ and that $P x = x - P_{M^\perp} x$ is the orthogonal projection, with $\|P\| = 1$ if $M \neq \{0\}$.

Problem 2. Show that $T^{-1}$ is bounded whenever $T$ is a bijective bounded operator between Banach spaces. Do not cite the open mapping theorem; prove it, and identify where completeness of both spaces is used.

Problem 3. Let $f_n \in X^*$ and suppose $\lim_n f_n(x)$ exists for every $x \in X$. Show the limit functional is bounded. Which theorem applies, and what would fail if $X$ were only a normed space?

Problem 4. Show the differentiation operator $D : C^1[0,1] \to C[0,1]$, $Df = f'$ is unbounded when $C^1[0,1]$ carries $\|\cdot\|_\infty$, and bounded when it carries $\|f\|_\infty + \|f'\|_\infty$. What does this say about why a domain is part of an operator’s definition?

Problem 5. Prove that if $T \in \mathcal{B}(X, X)$ with $\|T\| < 1$ then $I - T$ is invertible, constructing the inverse explicitly.

Problem 6. Show the closed graph theorem follows from the open mapping theorem and the bounded inverse theorem.

Problem 7. Prove that if $T : X \to Y$ is linear between Banach spaces and $x_n \to 0$ with $Tx_n \to y$ forces $y = 0$, then $T$ is bounded.

Problem 8 (important). Someone claims: “for any closed subspace $M$ of a Banach space $X$, the quotient map $\pi : X \to X/M$ has an isometric right inverse.” Find the flaw. The claim is true when $X$ is a Hilbert space (Problem 1), so the argument is to identify exactly which step uses Hilbert structure, and then to give an example of a Banach space with a closed subspace admitting no complement at all.

• All three Baire-based theorems are one theorem applied three times, each following the pattern: assume a pathology, cover the space by level sets of a bound, get interior from Baire, upgrade pointwise to uniform.
• Completeness is necessary and not decorative. $\mathbb{Q}$ is meagre in itself, so Baire fails there and so does the open mapping theorem.
• Hahn-Banach needs neither Baire nor completeness, and is the only one of the four that holds on a bare vector space with a sublinear gauge.
• Closed graph follows from open mapping by completing the domain, then showing the domain was everything.
• A bijective bounded operator between Banach spaces has a bounded inverse. Drop completeness on either side and this fails, which is the standard way to exhibit incomplete spaces.
• One norm dominating another is not equivalence unless both norms make the space Banach.

• Confusing Hahn-Banach with the open mapping theorem. Hahn-Banach extends functionals from subspaces while preserving the norm; the open mapping theorem is about surjectivity of operators and openness of images. Extension versus mapping. Substituting one for the other is the most common confusion in the subject.
• Applying the open mapping theorem without checking hypotheses. Both spaces must be Banach and $T$ must be surjective. The identity from an incomplete normed space to its completion is bijective and continuous but not open.
• Reading uniform boundedness as pointwise boundedness. UBP’s hypothesis is $\sup_\alpha \|T_\alpha x\| < \infty$ for each $x$, with the bound allowed to depend on $x$; the conclusion is a bound independent of $x$. That gap is the content of the theorem, and it needs $X$ Banach.
• Assuming one norm inequality gives equivalence. It gives continuity in one direction only. Equivalence needs both, and completeness is what manufactures the second from the first.
• Claiming every closed subspace of a Banach space has a complement. True for Hilbert spaces via orthogonal complements; false in general. See Problem 8.
• Proving the closed graph theorem by recursion on the graph. It is circular: the recursive step needs the theorem. Complete the domain and use openness, then argue the domain was everything.
• Believing completeness is technical. Every Baire-based theorem above is false on an incomplete space. $\mathbb{Q}$ shows the mechanism in one line.

• Normed Spaces and Banach Spaces — completeness, and the fact that finite-dimensional spaces are automatically complete.
• Bounded Linear Operators — $\mathcal{B}(X,Y)$, operator norm, and the unbounded operators these theorems exclude.
• Compact Operators — the Fredholm alternative, which uses the bounded inverse theorem.
• Weak and Weak* Convergence — Banach-Alaoglu, a weak*-compactness statement, and the weak/norm boundedness contrast from uniform boundedness.
• Sequences and Limits — Cauchy sequences and completeness.
• Homomorphisms and Isomorphism Theorems — quotients, which is what $X/M$ in Problem 8 is.
• Classical Mechanics

• Applications
• Bounded Linear Operators
• Compact Operators
• Historical Context