The open mapping theorem

Theorem

The open mapping theorem

T-096Home MU-304Threads space · structure

Statement

Let \(X\) and \(Y\) be Banach spaces over \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T : X \to Y\) be a linear operator that is bounded (i.e. \(T \in \mathcal{B}(X,Y)\)) and surjective, \(T(X) = Y\). Then \(T\) is an open map: for every open set \(U \subseteq X\), the image \(T(U)\) is open in \(Y\). Equivalently, there exists \(r \gt 0\) such that \(T(B_X(0,1)) \supseteq B_Y(0,r)\), where \(B_X(0,1) = \{x \in X : \|x\| \lt 1\}\).

Why it matters

The open mapping theorem is one of the three pillars of linear functional analysis alongside the uniform boundedness principle and the Hahn–Banach theorem. It converts a purely set-theoretic/algebraic fact (surjectivity) into a topological/metric fact (openness), and this upgrade is precisely what completeness of both spaces buys you. Its immediate corollary, the bounded inverse theorem, is the standard tool for proving that a continuous linear bijection between Banach spaces automatically has a continuous inverse — a fact used constantly to establish equivalence of norms and to certify that quotient maps \(X \to X/M\) are topological quotient maps.

It is also the engine behind the closed graph theorem (via the graph norm trick) and underlies why "define an operator and check it is bounded and bijective" is enough to get a Banach space isomorphism, without ever having to estimate the inverse directly.

Hypotheses

\(X\) is complete (a Banach space). Let \(X_0 = c_{00}\), the finitely supported sequences, with the \(\ell^1\) norm (incomplete). Define \(T : X_0 \to X_0\) by \((Tx)_n = x_n/n\). Then \(T\) is bounded, linear, bijective (given finitely supported \(y\), \(x_n = n y_n\) is again finitely supported), yet \(T^{-1}e_n = n e_n\) has \(\|T^{-1}e_n\| = n \to \infty\), so \(T^{-1}\) is unbounded and \(T\) is not open. Completeness of the domain is what the Baire category argument needs.

\(T\) is bounded (continuous). Take \(X=Y=\ell^2\) and let \(T\) be a linear bijection built from a Hamel basis that permutes basis vectors in a way that is unbounded (such maps exist assuming the axiom of choice, since a Hamel basis of an infinite-dimensional Banach space is uncountable and unbounded rearrangements of it are linear bijections). Being discontinuous, \(T\) is not even continuous, let alone open in the metric sense the theorem asserts; the theorem is vacuous without boundedness because openness is being deduced from metric estimates on \(\|Tx\|\) controlled by \(\|x\|\).

\(T\) is surjective. Take \(X = Y = \ell^2\) and let \(T\) be the (bounded, injective, non-surjective) right shift \(T(x_1,x_2,\dots) = (0,x_1,x_2,\dots)\). Then \(T(B_X(0,1))\) is contained in the hyperplane-like slab \(\{y : y_1 = 0\}\)-adjacent ball and contains no ball \(B_Y(0,r)\) for any \(r \gt 0\) around any interior point, because every image point has first coordinate \(0\); \(T\) is far from open. Surjectivity is what lets the Baire category argument exhaust all of \(Y\).

Proof

\text{Since } T \text{ is surjective}, \qquad Y = \bigcup_{n=1}^{\infty} T\big(n B_X(0,1)\big) = \bigcup_{n=1}^{\infty} n\, T\big(B_X(0,1)\big).

Every \(y \in Y\) is \(Ty\)-image of some \(x \in X\), and \(x \in nB_X(0,1)\) for \(n \gt \|x\|\); linearity gives \(T(nB_X(0,1)) = nT(B_X(0,1))\). A

Y = \bigcup_{n=1}^{\infty} \overline{n\,T(B_X(0,1))} \quad \text{and } Y \text{ is a complete metric space.}

Enlarging each set to its closure only makes the union bigger, so the identity of step 1 still gives a countable cover of \(Y\) by closed sets; \(Y\) is complete because it is a Banach space, hence a complete metric space in its norm metric. A

\text{By the Baire Category Theorem, some } \overline{n_0\, T(B_X(0,1))} \text{ has nonempty interior.}

Baire Category Theorem: a complete metric space is not a countable union of nowhere-dense closed sets. If every set in the cover of step 2 were nowhere dense, \(Y\) itself would be meagre in itself, contradicting completeness. So at least one has nonempty interior. B

\overline{T(B_X(0,1))} \text{ has nonempty interior.}

The map \(y \mapsto n_0 y\) is a homeomorphism of \(Y\) onto itself (linear, bounded, with bounded inverse \(y \mapsto y/n_0\)), and it carries \(\overline{T(B_X(0,1))}\) onto \(\overline{n_0T(B_X(0,1))}\); a homeomorphism preserves "having nonempty interior". A

\exists\, y_0 \in Y,\ \rho \gt 0 : \ B_Y(y_0,\rho) \subseteq \overline{T(B_X(0,1))} \ \implies\ B_Y(0,\rho) \subseteq \overline{T(B_X(0,1))}.

Let \(C = \overline{T(B_X(0,1))}\), the closure of the image of a symmetric convex set (the ball) under a linear map, hence \(C\) is itself convex and symmetric (\(C = -C\)) and closed. Given \(B(y_0,\rho)\subseteq C\), symmetry gives \(B(-y_0,\rho) \subseteq C\); convexity gives, for any \(z \in B(0,\rho)\), that \(z = \tfrac12(y_0+z) + \tfrac12(-y_0+z)\) is a midpoint of two points of \(C\) (since \(y_0+z \in B(y_0,\rho)\) and \(-y_0+z \in B(-y_0,\rho)\)), so \(z \in C\). C

\text{Rescaling: } \forall \varepsilon \gt 0,\qquad \overline{T(B_X(0,\varepsilon))} \supseteq B_Y(0,\rho\varepsilon).

Apply the bounded linear homeomorphism \(x \mapsto \varepsilon x\) to step 5's inclusion \(B_Y(0,\rho)\subseteq \overline{T(B_X(0,1))}\) and use linearity of \(T\), \(T(\varepsilon B_X(0,1)) = \varepsilon T(B_X(0,1))\). A

\textbf{Claim: } T(B_X(0,2)) \supseteq B_Y(0,\rho).

This is the key step that removes the closure bar left over from Baire category: we must show the (uncompleted) image genuinely contains a ball, not merely its closure. Proved in step 8 by successive approximation. B

\text{Fix } y \in B_Y(0,\rho). \text{ Build } x_1,x_2,\dots \text{ with } \|x_k\| \lt 2^{-(k-1)},\quad \Big\| y - \sum_{i=1}^{k} Tx_i \Big\| \lt \rho\, 2^{-k}.

By step 6 with \(\varepsilon=1\), \(\overline{T(B_X(0,1))}\supseteq B_Y(0,\rho)\), so choose \(x_1\), \(\|x_1\|\lt 1\), with \(\|y - Tx_1\|\lt \rho/2\). Inductively, having chosen \(x_1,\dots,x_k\) with residual \(y_k = y-\sum_{i\le k}Tx_i \in B_Y(0,\rho 2^{-k})\), apply step 6 with \(\varepsilon = 2^{-k}\) (which gives \(\overline{T(B_X(0,2^{-k}))}\supseteq B_Y(0,\rho 2^{-k})\)) to find \(x_{k+1}\), \(\|x_{k+1}\|\lt 2^{-k}\), with \(\|y_k - Tx_{k+1}\| \lt \rho 2^{-k-1}\). C

x := \sum_{i=1}^{\infty} x_i \text{ converges in } X, \qquad \|x\| \lt 2, \qquad Tx = y.

\(\sum \|x_i\| \lt \sum_{i\ge1} 2^{-(i-1)} = 2\), so the series is absolutely convergent; since \(X\) is complete (Banach space), absolute convergence implies convergence of the series to some \(x\in X\) with \(\|x\|\le \sum\|x_i\| \lt 2\), by the standard criterion for completeness via absolutely convergent series. Continuity (boundedness) of \(T\) lets us pass to the limit: \(Tx = \lim_k \sum_{i\le k}Tx_i = y\), because \(\|y-\sum_{i\le k}Tx_i\|\to 0\) by construction. C

\text{Hence } y \in T(B_X(0,2)) \text{ for every } y\in B_Y(0,\rho); \text{ set } r:=\rho/2 \text{ so } T(B_X(0,1)) \supseteq B_Y(0,r).

Step 9 shows every \(y\) with \(\|y\|\lt\rho\) is \(Tx\) for some \(x\) with \(\|x\|\lt2\); rescale by \(\tfrac12\) using linearity to land back in the unit ball. This proves Claim 7/the target radius statement with no closure remaining. A

T \text{ is an open map.}

Let \(U\subseteq X\) be open and \(y=Tx_0\in T(U)\) with \(x_0\in U\). Choose \(\varepsilon\gt0\) with \(x_0+B_X(0,\varepsilon)\subseteq U\) (definition of open set). Then \(T(U)\supseteq T(x_0) + T(B_X(0,\varepsilon)) \supseteq y + B_Y(0,r\varepsilon)\) by step 10 rescaled (linearity) using step 6's pattern, which is an open ball around \(y\). As \(y\in T(U)\) was arbitrary, \(T(U)\) is open. A

Result

T\in\mathcal{B}(X,Y),\ X,Y \text{ Banach},\ T(X)=Y \ \implies\ T \text{ is an open map}

Reading. A bounded linear surjection between complete normed spaces cannot be "topologically thin": it must spread every open set out into an open set in the target. Concretely, the image of the unit ball contains a fixed-radius ball around \(0\), so \(T\) cannot squeeze arbitrarily small pieces of \(Y\) out of arbitrarily large pieces of \(X\).

Scope. Requires both spaces complete and the map linear, bounded, and surjective; fails if any one of these four hypotheses is dropped (see counterexamples above and below). It holds equally for real and complex scalars and does not require \(T\) injective.

Corollaries & converses

Bounded Inverse Theorem. If \(T\in\mathcal{B}(X,Y)\) is a bijection between Banach spaces, then \(T^{-1}\) is automatically bounded. (Apply openness: \(T^{-1}\) is continuous because preimages under \(T^{-1}\) of open sets are exactly images under \(T\) of open sets, which are open.)
Equivalence of comparable norms. If \(\|\cdot\|_1,\|\cdot\|_2\) are two norms on the same space making it Banach in each, and \(\|\cdot\|_1 \le C\|\cdot\|_2\) for some \(C\), then the norms are equivalent (apply the bounded inverse theorem to the identity map).
Quotient map is an isomorphism onto image. If \(M\subseteq X\) is closed, the induced bounded bijection \(X/M \to T(X)\) coming from a bounded \(T\) with kernel \(M\) is a topological isomorphism whenever \(T(X)\) is closed (so that it is itself Banach).
Converse, in the following precise sense, is true: a bounded linear map \(T: X\to Y\) between Banach spaces is open if and only if it is surjective. ("Only if" needs no completeness: if \(T(B_X(0,1))\) contains a ball \(B_Y(0,r)\), then since \(T(X)\) is a linear subspace containing a neighbourhood of \(0\), scaling shows \(T(X)=Y\).) So surjectivity and openness are equivalent for bounded operators between Banach spaces — the theorem's real content is the nontrivial implication surjective \(\Rightarrow\) open.
Does not converse to injectivity. Openness/surjectivity says nothing about injectivity: \(T\) can be open and have a large kernel (e.g. any bounded surjection with nontrivial kernel, such as a coordinate projection \(\ell^2\to\ell^2\), \((x_n)\mapsto(x_2,x_3,\dots)\)).

Fails without

\(X\) incomplete: \(T:c_{00}\to c_{00}\) (both with \(\ell^1\) norm), \((Tx)_n=x_n/n\), is a bounded bijection with unbounded inverse \(T^{-1}e_n=ne_n\); not open.
\(Y\) incomplete: the identity map \(\ell^1\to(\ell^1,\|\cdot\|_2)\) with \(\|y\|_2=\sum|y_n|/n\) a strictly weaker, incomplete norm on the same vector space; bounded bijection, not open because the norms are inequivalent.
\(T\) not surjective: the right shift \(T(x_1,x_2,\dots)=(0,x_1,x_2,\dots)\) on \(\ell^2\) is bounded and injective but its image, and hence \(T(B(0,1))\), contains no ball of \(\ell^2\) since every image point has vanishing first coordinate.
\(T\) unbounded (discontinuous linear bijection): a Hamel-basis-rearranging linear bijection of an infinite-dimensional Banach space onto itself need not be continuous, so it is not open in any sense controlled by the norm topology — the very statement "open map" loses its analytic content without boundedness feeding into the Baire argument.

Common errors

Forgetting to remove the closure bar (step 8–10): students often stop once they've shown \(\overline{T(B_X(0,1))}\supseteq B_Y(0,r)\) via Baire category and think this already proves openness — it only proves the closure of the image contains a ball, not the image itself.
Applying the theorem to non-surjective operators and expecting openness (e.g. assuming every bounded injective operator between Banach spaces has closed, let alone open, range).
Confusing "open map" with "maps open sets to sets containing an open set around each point of the image only when restricted to the unit ball" — openness must hold for every open \(U\), which needs the translation-and-rescale argument of step 11, not just the single statement about \(B_X(0,1)\).
Trying to prove the theorem for normed (not Banach) spaces by mimicking the argument without checking Baire's completeness hypothesis on \(Y\), or without checking absolute convergence needs completeness of \(X\) (step 9).
Misapplying the bounded inverse theorem corollary to non-bijective maps, or to bijective maps between spaces where one of the two is not complete.

Discussion

The open mapping theorem, together with the uniform boundedness principle and the closed graph theorem, forms the trio of "big three" consequences of the Baire category theorem in functional analysis; all three exploit completeness in an essentially non-constructive way — the theorem asserts existence of the radius \(r\) but the proof gives no explicit value of \(r\) in terms of \(T\) beyond what Baire category supplies. This is characteristic: category arguments prove existence via contradiction with meagreness, not by exhibiting witnesses.

Historically the theorem originates with Juliusz Schauder's 1930 refinement of ideas from Stefan Banach's foundational work on normed spaces, and it is a direct descendant of Baire's 1899 category theorem for complete metric spaces, itself motivated by real-analysis questions about the size of exceptional sets. The theorem is a striking example of "soft analysis": it converts information about how big \(T(X)\) is set-theoretically (all of \(Y\)) into a quantitative estimate (a fixed \(r\) works uniformly), purely from completeness, without any structure specific to Hilbert spaces or reflexivity.

A useful structural viewpoint: the theorem says that a bounded linear surjection \(T:X\to Y\) between Banach spaces factors as \(X \xrightarrow{\pi} X/\ker T \xrightarrow{\widetilde T} Y\) where \(\pi\) is the (always open) quotient map and \(\widetilde T\) is a bounded linear bijection; the open mapping theorem is exactly the statement that this induced \(\widetilde T\) is a topological isomorphism, i.e. \(Y \cong X/\ker T\) as Banach spaces, not merely as vector spaces.

A subtlety worth flagging: the theorem is genuinely about Banach (or more generally Fréchet) spaces; in general topological vector spaces that are not Baire spaces, surjective continuous linear maps need not be open, and the appropriate generalization (open mapping theorem for Fréchet spaces, or for F-spaces) still needs a Baire-type completeness hypothesis on the domain, with the same closure-removal argument going through verbatim once "ball" is replaced by a basic neighbourhood of \(0\) in a translation-invariant metric. Common misconception: that the theorem needs \(T\) injective, or that it is equivalent to the closed graph theorem outright — in fact the closed graph theorem is usually derived from the open mapping theorem (applied to the projection from the graph, itself Banach under the graph norm, onto \(X\)), not the reverse; the logical dependency runs one way in the standard development, though the two are ultimately mutually derivable using the closed range and bounded inverse machinery.

Worked examples

\text{Show } T:\ell^2\to\ell^2,\quad T(x_1,x_2,x_3,\dots) = (x_1,\, x_2/2,\, x_3/3+x_1,\, x_4,x_5,\dots)\ \text{needs justification before applying the theorem.}

Before invoking openness we must check the actual hypotheses hold; guessing "it's linear so it's fine" is exactly the error flagged above. A

\text{Instead take the concrete, checkable example } T:\ell^2\to\ell^2,\quad Tx = (x_2,x_3,x_4,\dots)\ \text{(the left shift, dropping the first coordinate).}

This is linear, and \(\|Tx\|_2 \le \|x\|_2\) so it is bounded with \(\|T\|\le1\); it is surjective because given any \(y=(y_1,y_2,\dots)\in\ell^2\), \(x=(0,y_1,y_2,\dots)\in\ell^2\) satisfies \(Tx=y\). A

\ell^2 \text{ is a Hilbert space, hence Banach; both hypotheses of the theorem (completeness of domain and codomain) hold.}

Standard fact: \(\ell^2\) with the norm \(\|x\|_2=(\sum|x_n|^2)^{1/2}\) is complete (a short application of the Riesz–Fischer style diagonal/Cauchy-in-each-coordinate argument). A

\text{By the theorem, } T \text{ is open; explicitly, taking } x=(0,y_1,y_2,\dots) \text{ shows } \|x\|=\|y\|,\ \text{so } T(B(0,1))\supseteq B(0,1).

Here we can beat the abstract proof and exhibit \(r=1\) directly: for \(\|y\|\lt1\), the explicit preimage \(x=(0,y_1,y_2,\dots)\) has \(\|x\|=\|y\|\lt1\), so \(y\in T(B(0,1))\) with no need for the Baire successive-approximation machinery, illustrating that the theorem's radius, while non-constructive in general, is sometimes computable by hand. B

T(x_1,x_2,\dots)=(x_2,x_3,\dots)\ \text{on}\ \ell^2\ \text{is a bounded, surjective, open map, with explicit radius } r=1.

Reading. The coordinate-drop operator is bounded, onto, and (as the theorem guarantees) topologically well-behaved: it never crushes an open set into something thin.

\text{Let } X=C[0,1]\ (\text{sup norm}),\ Y=\mathbb{R}^2,\ \text{and } T(f) = \big(f(0),\, f(1)\big).

Set-up: evaluate a continuous function at the two endpoints of \([0,1]\); this is manifestly linear in \(f\). A

T \text{ is bounded: } \|Tf\|_{\infty} = \max(|f(0)|,|f(1)|) \le \|f\|_{\sup}.

Direct estimate from the definition of the sup norm on \(C[0,1]\), giving \(\|T\|\le 1\). A

T \text{ is surjective: given } (a,b)\in\mathbb{R}^2, \text{ the affine function } f(t) = a+(b-a)t \text{ satisfies } Tf=(a,b).

Explicit witness of surjectivity, all that is required by the definition \(T(X)=Y\). A

X=C[0,1] \text{ with sup norm is Banach (uniform limit of continuous functions is continuous); } Y=\mathbb{R}^2 \text{ is Banach (finite-dimensional).}

Both completeness hypotheses of the theorem verified: the first is the classical uniform-limit theorem from real analysis, the second because all norms on \(\mathbb{R}^2\) are equivalent and \(\mathbb{R}^2\) is complete in the Euclidean metric. A

\text{By the theorem, } T \text{ is open; take } (a,b) \text{ with } \max(|a|,|b|)\lt \tfrac12 \text{ and exhibit } f \text{ with } \|f\|_{\sup}\lt 1, Tf=(a,b).

The affine function of step 3 has \(\|f\|_{\sup} = \max(|a|,|b|,|\text{value at any interior point}|)\); since \(f\) is affine its sup over \([0,1]\) is \(\max(|a|,|b|)\), so choosing \(\max(|a|,|b|)\lt1\) already gives \(\|f\|_{\sup}\lt 1\), witnessing \(T(B_X(0,1))\supseteq B_Y(0,1)\) with the sharp radius \(r=1\), consistent with (in fact stronger than) what the abstract theorem promises. B

T(f)=(f(0),f(1)):\ C[0,1]\to\mathbb{R}^2\ \text{is open, with}\ T(B_X(0,1))\supseteq B_Y(0,1).

Reading. Boundary-value evaluation on \(C[0,1]\) is a bounded surjection onto \(\mathbb{R}^2\) between Banach spaces, so the open mapping theorem applies; here the sharp constant can even be computed by hand via the affine interpolant.

Problems

Prove directly (without citing the open mapping theorem) that the map \(T:\ell^1\to\ell^1\), \(Tx = (x_1+x_2, x_2, x_3,\dots)\) is bounded and bijective, and find an explicit \(r\gt0\) with \(T(B(0,1))\supseteq B(0,r)\).
Solution
Boundedness: \(\|Tx\|_1 = |x_1+x_2| + \sum_{n\ge2}|x_n| \le |x_1|+2|x_2|+\sum_{n\ge3}|x_n| \le 2\|x\|_1\), so \(\|T\|\le2\). Injectivity: if \(Tx=0\) then \(x_2=0,x_3=0,\dots\) and \(x_1+x_2=0\Rightarrow x_1=0\), so \(x=0\). Surjectivity: given \(y\in\ell^1\), set \(x_2=y_2,x_3=y_3,\dots\) and \(x_1=y_1-y_2\); then \(x\in\ell^1\) (finite sum of \(\ell^1\) sequences) and \(Tx=y\) by direct check. Explicit radius: for \(\|y\|_1\lt r\), \(\|x\|_1 = |y_1-y_2|+\sum_{n\ge2}|y_n| \le 2\|y\|_1\), so \(\|x\|_1\lt 2r\); taking \(r=1/2\) gives \(\|x\|_1\lt1\), so \(T(B(0,1))\supseteq B(0,1/2)\).
Let \(X\) be a Banach space and \(M\subset X\) a proper closed subspace. Using the open mapping theorem, show the quotient map \(\pi:X\to X/M\) (with quotient norm \(\|x+M\| = \inf_{m\in M}\|x-m\|\)) is open.
Solution
\(X/M\) is a Banach space under the quotient norm (standard fact: quotient of Banach by closed subspace is Banach, itself provable via absolutely convergent series as in step 9). \(\pi\) is linear, and bounded with \(\|\pi\|\le1\) since \(\|\pi x\| = \inf_m\|x-m\|\le\|x-0\|=\|x\|\). \(\pi\) is surjective by definition of the quotient (every coset \(x+M\) is \(\pi(x)\)). Both \(X\) and \(X/M\) are Banach, \(\pi\) is bounded and surjective, so by the open mapping theorem \(\pi\) is open.
Give an example of a bounded linear bijection \(T\) between two Banach spaces where the sharp radius \(r\) in \(T(B(0,1))\supseteq B(0,r)\) equals \(\|T^{-1}\|^{-1}\), and prove this identity in general.
Solution
For any bounded bijection \(T:X\to Y\) between Banach spaces with bounded inverse (guaranteed by the theorem), the sharp \(r\) is exactly \(1/\|T^{-1}\|\): if \(\|y\|\lt r\) with \(r=1/\|T^{-1}\|\), then \(x=T^{-1}y\) satisfies \(\|x\|\le\|T^{-1}\|\,\|y\|\lt \|T^{-1}\|\cdot r=1\), so \(x\in B(0,1)\) and \(Tx=y\), giving \(T(B(0,1))\supseteq B(0,r)\); conversely no larger radius works because \(\|T^{-1}\| = \sup_{\|y\|=1}\|T^{-1}y\|\) means there exist \(y_n\), \(\|y_n\|=1\), with \(\|T^{-1}y_n\|\to\|T^{-1}\|\), so \(y_n/(\|T^{-1}\|+\epsilon)\) is not attainable from inside \(B(0,1)\) for radius bigger than \(r\). A concrete instance: \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x_1,x_2)=(2x_1,x_2)\), has \(T^{-1}(y_1,y_2)=(y_1/2,y_2)\), \(\|T^{-1}\|=1\) (operator norm, sup norm on \(\mathbb{R}^2\)), matching \(r=1\), and indeed \(T(B(0,1))\) is the rectangle \([-2,2]\times[-1,1]\) whose largest inscribed ball around 0 (sup norm) has radius \(1\).
Let \(X=C[0,1]\) with sup norm and let \(Y=C[0,1]\) with the (incomplete) \(L^1\)-norm \(\|f\|_1=\int_0^1|f|\). Show the identity map \(T:X\to Y\) is bounded and bijective (as a linear map on the same underlying set of functions) but not open, and explain why this does not contradict the open mapping theorem.
Solution
Boundedness: \(\|f\|_1 = \int_0^1|f| \le \|f\|_{\sup}\), so \(\|T\|\le1\). Bijectivity as a map of the underlying vector space \(C[0,1]\) is trivial (identity on functions). Not open: if it were open, \(T^{-1}\) would be bounded by the bounded inverse theorem corollary, i.e. \(\|f\|_{\sup}\le C\|f\|_1\) for all \(f\in C[0,1]\); but the "tent" functions \(f_n\) that spike to height \(1\) on an interval of width \(1/n\) around a point and are \(0\) elsewhere have \(\|f_n\|_{\sup}=1\) while \(\|f_n\|_1\to0\), so no such \(C\) exists. This does not contradict the theorem because its hypotheses require both \(X\) and \(Y\) complete; here \(Y=(C[0,1],\|\cdot\|_1)\) is not complete (its completion is \(L^1[0,1]\), a strictly larger space), so the theorem simply does not apply.
(Closed graph via open mapping.) Let \(X,Y\) be Banach spaces and \(T:X\to Y\) linear with closed graph \(\Gamma(T)=\{(x,Tx):x\in X\}\subseteq X\times Y\). Using the open mapping theorem, sketch why \(T\) must be bounded.
Solution
Equip \(X\times Y\) with the norm \(\|(x,y)\| = \|x\|_X+\|y\|_Y\), under which \(X\times Y\) is Banach (a Cauchy sequence of pairs is coordinatewise Cauchy, and each coordinate space is complete). Since \(\Gamma(T)\) is a closed subspace of a Banach space, \(\Gamma(T)\) is itself Banach (closed subspaces of complete spaces are complete). The projection \(p_1:\Gamma(T)\to X\), \((x,Tx)\mapsto x\), is linear, bounded (\(\|p_1(x,Tx)\|=\|x\|\le\|(x,Tx)\|\)), and bijective (inverse \(x\mapsto(x,Tx)\), well-defined since \(T\) is a function). By the open mapping theorem (equivalently the bounded inverse theorem corollary), \(p_1^{-1}:X\to\Gamma(T)\), \(x\mapsto(x,Tx)\), is bounded: \(\|(x,Tx)\|\le C\|x\|\) for some \(C\), i.e. \(\|x\|+\|Tx\|\le C\|x\|\), so \(\|Tx\|\le(C-1)\|x\|\), proving \(T\) is bounded.