The closed graph theorem

Statement

Let \(X\) and \(Y\) be Banach spaces over \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:X\to Y\) be a linear map (not assumed continuous). Define the graph of \(T\) to be \(G(T)=\{(x,Tx):x\in X\}\subseteq X\times Y\), where \(X\times Y\) carries the norm \(\|(x,y)\|=\|x\|_X+\|y\|_Y\) (or any equivalent product norm). If \(G(T)\) is a closed subset of \(X\times Y\), then \(T\) is bounded, i.e. \(T\in\mathcal{B}(X,Y)\) with \(\|T\|=\sup_{\|x\|\le 1}\|Tx\|_Y \lt \infty\).

Why it matters

Continuity of a linear map between normed spaces is usually checked by a direct \(\varepsilon\)-\(\delta\) (equivalently, sequential) argument: show \(x_n\to x\) forces \(Tx_n\to Tx\). The closed graph theorem replaces this with a strictly weaker-looking hypothesis: show that \(x_n\to x\) and \(Tx_n\to y\) together force \(y=Tx\). This is often far easier to verify — differential and integral operators, adjoints defined implicitly, and maps built from other closed operators are all naturally handled this way — and the theorem certifies that on Banach spaces the two notions coincide.

It is one of the three classical consequences of the Baire category theorem in functional analysis, alongside the open mapping theorem and the uniform boundedness principle, and in practice it is often the most directly applicable of the three, since verifying closedness of a graph is a completeness/limit argument with no need to produce an explicit bound.

Hypotheses

Proof

We give the standard proof via the open mapping theorem, using the graph norm on \(X\).

Result

Corollaries & converses

• Converse holds trivially: if \(T\) is bounded (continuous) then \(G(T)\) is always closed, for any normed spaces \(X,Y\) (no completeness needed) — if \(x_n\to x\) and \(Tx_n\to y\), continuity gives \(Tx_n\to Tx\), and limits in Hausdorff spaces are unique, so \(y=Tx\). So on Banach spaces "bounded" and "closed graph" are logically equivalent.
• If \(T:X\to Y\) is linear, everywhere-defined, and closed as an unbounded operator (graph closed) between Banach spaces, then automatically \(T\) has no genuinely unbounded behaviour to exhibit — this is the standard method for proving an operator defined by a seemingly delicate formula (e.g. via a limit, a series, or an integral) is in fact bounded.
• Corollary (uniform boundedness is not needed): unlike the uniform boundedness principle, the closed graph theorem requires no family of operators — it is a single-operator statement, making it typically the tool of choice when boundedness of one specific map is in question.
• If \(T\) is a linear bijection between Banach spaces with closed graph, \(T^{-1}\) also has closed graph (the graph of \(T^{-1}\) is the "flip" of the graph of \(T\), still closed), so \(T^{-1}\) is bounded too — recovering the open mapping theorem's corollary for bijections as a special case.

Fails without

Common errors

• Confusing "graph is closed" with "\(T\) is continuous" as if they were definitionally the same statement; students then skip the proof entirely. They are equivalent only because of completeness — the theorem is a nontrivial consequence of Baire category, not a tautology.
• Trying to verify closedness by checking continuity of \(T\) directly (i.e. assuming \(x_n\to x\) implies \(Tx_n\to Tx\)) — this begs the question. The correct check assumes \(x_n\to x\) and \(Tx_n\to y\) as two separate hypotheses, then must independently show \(y=Tx\), often via a different (e.g. distributional, or weak) argument that does not presuppose continuity.
• Forgetting the completeness hypotheses on \(X\) and \(Y\) and trying to apply the theorem to operators on incomplete normed spaces, e.g. spaces of polynomials with a sup norm, or \(C^1\) functions with the \(C^0\) norm.
• Applying the closed graph theorem to unbounded, densely-defined operators (e.g. \(\frac{d}{dx}\) on \(D(T)=C^1[0,1]\subset C[0,1]\)) and concluding they are bounded — the theorem requires \(T\) to be defined on all of \(X\), which fails here; the graph can be closed while the operator is genuinely unbounded on the smaller domain.
• Misremembering which direction needs completeness: the converse (bounded \(\Rightarrow\) closed graph) needs no completeness at all and holds in any normed space; only the forward direction needs both spaces Banach.

Discussion

The closed graph theorem, the open mapping theorem, and the uniform boundedness principle are the three pillars built on the Baire category theorem in functional analysis, and they are not independent: each can be derived from either of the others with modest extra work. The proof given here derives the closed graph theorem from the open mapping theorem via the graph-norm trick; historically Banach's 1932 Théorie des Opérations Linéaires established both results together as part of his systematic development of what are now called Banach spaces.

The graph-norm construction in Steps 1–3 is worth internalizing as a technique in its own right: whenever one has a linear map \(T\) whose graph is closed, the domain can be re-normed by \(\|x\|+\|Tx\|\) to make \(T\) "visibly" bounded (bounded by definition, with constant 1, with respect to the new norm), and the entire content of the theorem is that this new norm is actually equivalent to the old one — a fact that is utterly false in general (renorming a space can produce a strictly finer, inequivalent topology) and holds here only because completeness pins the two topologies together via Baire category.

In practice, the theorem is the workhorse for proving that operators defined by formulas involving limits, sums, or PDE data are bounded: one shows the graph is closed by a routine limit-interchange argument (often just "pass to a subsequence and use uniqueness of limits" or a distributional/weak argument), which is typically far more tractable than bounding \(\|Tx\|\) by \(\|x\|\) directly. Classic applications include showing that a symmetric operator on a Hilbert space with everywhere-defined adjoint is bounded (Hellinger–Toeplitz theorem, itself a corollary), and showing that certain naturally-defined maps between sequence or function spaces are automatically continuous.

Common misconception: that the closed graph theorem lets us conclude boundedness for any "reasonable-looking" linear operator. It emphatically does not apply to densely-defined unbounded operators central to spectral theory and PDE (e.g. the position and momentum operators in quantum mechanics, or elliptic differential operators) — these are typically closed operators on a proper dense domain \(D(T)\subsetneq X\), and it is precisely because \(D(T)\ne X\) that the theorem's conclusion is evaded; closedness there is a substitute for, not a route to, boundedness. The graph is closed with respect to the graph norm restricted to \(D(T)\), which need not make \((D(T),\|\cdot\|_G)\) coincide with \((X,\|\cdot\|_X)\) topologically, so Step 5 above has no analogue.

Worked examples

Problems

1. Let \(X=Y=\ell^2\) and let \(T\) be defined by \((Tx)_n = n x_n\) for \(x=(x_n)\in \ell^2\), with domain restricted to those \(x\) for which \((nx_n)\in\ell^2\), i.e. \(D(T)=\{x\in\ell^2 : \sum n^2|x_n|^2\lt\infty\}\ne \ell^2\). Explain precisely why the closed graph theorem does not force \(T\) to be bounded, even though one can check \(G(T)\) is closed in \(\ell^2\times \ell^2\).
   Solution

   The closed graph theorem requires \(T\) to be defined on the whole Banach space \(X\); here \(D(T)\subsetneq \ell^2\) is a proper dense subspace (it is dense since it contains all finitely-supported sequences), not all of \(\ell^2\). The hypothesis "\(T:X\to Y\) linear" in the theorem statement means \(X\) itself is the domain. Since \(T\) is only densely defined, the theorem's hypotheses are not met, and indeed the conclusion fails: \(T\) is unbounded on \(D(T)\) (take \(x=e_n\), \(\|e_n\|=1\), \(\|Te_n\|=n\to\infty\)), while \(G(T)\) is closed as a subset of \(\ell^2\times\ell^2\) (a standard check: if \(x^{(k)}\to x\) in \(\ell^2\) with \(x^{(k)}\in D(T)\) and \(Tx^{(k)}\to y\) in \(\ell^2\), then coordinatewise \(y_n = nx_n\), and one shows \(x\in D(T)\) with \(Tx=y\) using Fatou's lemma on \(\sum n^2|x_n|^2\)). This is the standard example of a closed, densely-defined, unbounded operator, and it shows the "everywhere-defined" hypothesis is not a technicality.
2. Let \(X\) be a Banach space and let \(T:X\to X\) be linear and idempotent (\(T^2=T\)) with closed graph. Show \(T\) is bounded, and that \(X = \ker T \oplus \operatorname{ran} T\) as a topological direct sum (both summands closed).
   Solution

   Boundedness is immediate from the closed graph theorem: \(X\) is Banach, \(T\) linear, graph closed, so \(T\in\mathcal{B}(X)\). For the direct sum: \(\ker T\) is closed since \(T\) is continuous (preimage of the closed set \(\{0\}\)). \(\operatorname{ran}T = \ker(I-T)\): indeed if \(y=Tx\) then \((I-T)y = Tx - T^2x = Tx-Tx=0\); conversely if \((I-T)y=0\) then \(y=Ty\in\operatorname{ran}T\). Since \(I-T\) is also bounded (as \(T\) is), \(\ker(I-T)\) is closed, so \(\operatorname{ran}T\) is closed. Every \(x\in X\) decomposes as \(x = Tx + (I-T)x\), with \(Tx\in\operatorname{ran}T\) and \((I-T)x\in\ker T\) (since \(T(I-T)x = Tx-T^2x=0\)); uniqueness of the decomposition follows because \(\ker T\cap\operatorname{ran}T=\{0\}\) (if \(z\in\ker T\) and \(z=Tw\), then \(z=Tw=T^2w=Tz=0\)). The projections \(T\) and \(I-T\) onto the two closed subspaces are bounded, which is exactly what "topological direct sum" requires.
3. Give a complete proof that the converse direction of the closed graph theorem — "\(T\) bounded \(\Rightarrow\) \(G(T)\) closed" — holds for linear \(T\) between *arbitrary* normed spaces \(X,Y\) (no completeness assumed), and identify exactly where completeness would have been needed had you tried to prove the forward direction with the same method.
   Solution

   Suppose \(T\in\mathcal{B}(X,Y)\), and let \((x_n,Tx_n)\to (x,y)\) in \(X\times Y\), i.e. \(x_n\to x\) in \(X\) and \(Tx_n\to y\) in \(Y\). Continuity of \(T\) gives \(Tx_n\to Tx\) in \(Y\) (since \(\|Tx_n-Tx\|\le\|T\|\,\|x_n-x\|\to0\)). But \(Tx_n\to y\) also, and limits in a normed space (a Hausdorff topological space) are unique, so \(y=Tx\). Hence \((x,y)=(x,Tx)\in G(T)\), so \(G(T)\) is closed — no completeness of \(X\) or \(Y\) was used anywhere in this argument, only continuity of \(T\) and uniqueness of limits. Had one instead tried to prove the forward direction (closed graph \(\Rightarrow\) bounded) without completeness, the argument breaks at the step invoking the open mapping theorem (our Step 5), which itself is proved via the Baire category theorem applied to \(X\times Y\); Baire category requires a complete (or locally compact) space, so both \(X\) and \(Y\) Banach is essential there, with no analogous route available in incomplete spaces (indeed Question/Hypothesis counterexamples above show the forward implication genuinely fails without completeness).
4. Let \(X\) be a Banach space and \(Y\) a Banach space, and suppose \(T:X\to Y\) is linear with the weaker property that for every \(f\in Y^*\) (continuous linear functional), the composite \(f\circ T : X\to \mathbb{K}\) is bounded. Does it follow that \(T\) itself is bounded? Prove it using the closed graph theorem, citing the Hahn–Banach theorem where needed.
   Solution

   Yes. We show \(G(T)\) is closed. Let \(x_n\to x\) in \(X\) with \(Tx_n\to y\) in \(Y\); we must show \(y=Tx\). For any \(f\in Y^*\), \(f\circ T\) is bounded (given) hence continuous, so \(f(Tx_n)\to f(Tx)\) (continuity of \(f\circ T\) at \(x\), since \(x_n\to x\)). On the other hand, \(Tx_n\to y\) in \(Y\) and \(f\) is continuous, so \(f(Tx_n)\to f(y)\). By uniqueness of limits in \(\mathbb{K}\), \(f(Tx)=f(y)\) for every \(f\in Y^*\). By the Hahn–Banach theorem (specifically its corollary that \(Y^*\) separates points of \(Y\): if \(f(u)=f(v)\) for all \(f\in Y^*\) then \(u=v\), obtained by applying Hahn–Banach to extend a norming functional for \(u-v\)), we conclude \(Tx=y\). Hence \(G(T)\) is closed, \(X,Y\) are Banach, \(T\) is linear, so the closed graph theorem gives \(T\in\mathcal{B}(X,Y)\). (This also reproves that weak and strong boundedness coincide for linear operators between Banach spaces.)
5. (Converse-type sharpening) Let \(X,Y\) be Banach spaces and \(T:X\to Y\) linear and bounded. Let \(Z\subseteq X\) be a closed subspace. Show directly, without re-invoking the closed graph theorem, that \(T|_Z:Z\to Y\) is bounded, and then explain why this is not a genuinely new instance of the closed graph theorem but rather a triviality — contrast this with a scenario where restricting a *closed-graph but unbounded-looking* operator to a subspace is the substantive step.
   Solution

   Directly: \(Z\) is a closed subspace of the Banach space \(X\), hence \((Z,\|\cdot\|_X)\) is itself a Banach space (closed subspaces of complete spaces are complete). For \(z\in Z\), \(\|T|_Z(z)\|_Y = \|Tz\|_Y \le \|T\|\,\|z\|_X\), so \(T|_Z\) is bounded with \(\|T|_Z\|\le\|T\|\) — this uses only that \(T\) was already bounded on all of \(X\), so restricting to a subspace trivially preserves the same bound; no closed graph theorem is needed, since we already have an explicit constant. The theorem becomes substantive in the reverse-flavoured situation: suppose instead we only know \(S:X\to Y\) is linear with closed graph (not yet known bounded), and we want to bound \(S\) on a closed subspace \(Z\). One route is to check that \(G(S|_Z) = G(S)\cap (Z\times Y)\) is closed in \(Z\times Y\) (intersection of closed sets is closed, and \(Z\times Y\) is closed in \(X\times Y\)), then apply the closed graph theorem to \(S|_Z:Z\to Y\) directly, using completeness of \(Z\) — this recovers boundedness of \(S\) on \(Z\) without first establishing it on all of \(X\), which is useful when \(S\) is only known to have closed graph on the smaller space, not on all of \(X\).