From Openness to Closed Graphs
The Open Mapping Theorem concerns bounded surjective linear maps between Banach spaces. A closely related result starts with a different kind of information: instead of assuming an operator is bounded, we assume its graph is closed. Under the right hypotheses, closedness forces boundedness. This is useful when a formula defines an operator but a direct norm estimate is difficult to find.
The connection with the previous tutorial is through a projection. The graph of an operator is a subspace of a product space. If that graph is closed, it is itself a Banach space. Projecting the graph onto the domain gives a bounded linear bijection, so the Bounded Inverse Theorem obtained from the Open Mapping Theorem controls the inverse projection. That control yields a bound for the original operator.
For a linear map between normed spaces, boundedness is equivalent to continuity. The Closed Graph Theorem therefore says that an everywhere-defined linear map between Banach spaces is continuous if its graph is closed. The condition “everywhere-defined” matters: a closed operator whose domain is only a proper subspace need not be bounded with respect to the norm inherited from the ambient space.
A Sequential Test for Closedness
Because \(X\times Y\) is a metric space, closedness of the graph can be checked by limits of sequences. The following criterion translates the geometric definition into a practical test.
Proof. Suppose first that \(\Gamma(T)\) is closed. If \(x_n\to x\) and \(Tx_n\to y\), then \((x_n,Tx_n)\to(x,y)\) in the product norm, since \(\max(\|x_n-x\|_X,\|Tx_n-y\|_Y)\to0\). Each \((x_n,Tx_n)\) belongs to \(\Gamma(T)\). Closedness therefore gives \((x,y)\in\Gamma(T)\), which means \(y=Tx\).
Conversely, suppose the stated sequential condition holds. Let \((x,y)\) be in the closure of \(\Gamma(T)\). In a metric space, for each positive integer \(n\), we can choose \((x_n,Tx_n)\in\Gamma(T)\) whose distance from \((x,y)\) is less than \(1/n\). Thus \(x_n\to x\) and \(Tx_n\to y\). The condition gives \(y=Tx\), so \((x,y)\in\Gamma(T)\). Hence the graph contains all its closure points and is closed. \(\square\)
The test requires both limits. Convergence of \(x_n\) alone tells us nothing about whether the images converge, and convergence of the images alone does not identify their input. When both limits exist, closedness says their values must fit the operator’s defining relation.
The Closed Graph Theorem
Proof. Equip \(X\times Y\) with the maximum norm. This product is Banach: if \(((x_n,y_n))\) is Cauchy in that norm, then \((x_n)\) is Cauchy in \(X\) and \((y_n)\) is Cauchy in \(Y\). Completeness gives limits \(x\in X\) and \(y\in Y\), and then \((x_n,y_n)\to(x,y)\) in the product norm.
Since \(\Gamma(T)\) is closed in the Banach space \(X\times Y\), it is Banach with the inherited norm. Define the first-coordinate projection \(P_X:\Gamma(T)\to X\) by \(P_X(x,Tx)=x\). It is linear and bounded, because \(\|P_X(x,Tx)\|_X=\|x\|_X\leq\max(\|x\|_X,\|Tx\|_Y)\). It is surjective because \(T\) is defined on all of \(X\), and it is injective because a graph contains exactly one point with first coordinate \(x\). Thus \(P_X\) is a bounded linear bijection between Banach spaces.
By the Bounded Inverse Theorem, \((P_X)^{-1}:X\to\Gamma(T)\) is bounded. The second-coordinate projection \(P_Y:\Gamma(T)\to Y\), given by \(P_Y(x,Tx)=Tx\), is also bounded, since \(\|P_Y(x,Tx)\|_Y\leq\max(\|x\|_X,\|Tx\|_Y)\). For every \(x\in X\), \(T x=P_Y((P_X)^{-1}x)\). Therefore \(T=P_Y\circ(P_X)^{-1}\) is bounded as a composition of bounded linear maps. \(\square\)
This proof does not estimate \(Tx\) directly from the formula for \(T\). Instead, it puts the graph into a complete space and uses the Open Mapping Theorem through the bounded inverse result. The key structural fact is that the projection from the graph to the domain is a bijection precisely because the operator is defined everywhere and has a unique output at each input.
Worked Examples
Worked Example: Closedness Gives a Bound for an Integral Operator
Let \(X=Y=C([0,1])\), with the supremum norm, and define \((Tf)(x)=\int_0^x (1+t)f(t)\,dt\). The integral defines a continuous function: its integrand is continuous, and the fundamental theorem of calculus gives \((Tf)'(x)=(1+x)f(x)\). The map is linear.
To verify that its graph is closed, suppose \(f_n\to f\) uniformly and \(Tf_n\to g\) uniformly. For every \(x\in[0,1]\), \[ |(Tf_n)(x)-(Tf)(x)| \leq \|f_n-f\|_\infty\int_0^x(1+t)\,dt \leq \frac{3}{2}\|f_n-f\|_\infty. \] Thus \(Tf_n\to Tf\) uniformly. Since also \(Tf_n\to g\) uniformly, uniqueness of limits gives \(g=Tf\). The sequential criterion shows that the graph is closed. Both spaces are Banach, so the Closed Graph Theorem implies that \(T\) is bounded.
Here a direct estimate is also available: \(\|Tf\|_\infty\leq\frac{3}{2}\|f\|_\infty\). The example illustrates the closed-graph method: identify the limit of the outputs from the formula, then invoke the theorem for boundedness.
Worked Example: Composition with a Continuous Change of Variable
Define \(T:C([0,1])\to C([0,1])\) by \((Tf)(x)=f(x^2)\). The map \(x\mapsto x^2\) sends \([0,1]\) into itself, so \(Tf\) is continuous whenever \(f\) is continuous. The operator is linear.
Suppose \(f_n\to f\) uniformly and \(Tf_n\to g\) uniformly. For each \(x\in[0,1]\), \[ |(Tf_n)(x)-(Tf)(x)|=|f_n(x^2)-f(x^2)|\leq\|f_n-f\|_\infty. \] It follows that \(Tf_n\to Tf\) uniformly. Uniqueness of uniform limits gives \(g=Tf\), so the graph is closed. The Closed Graph Theorem applies because \(C([0,1])\) is Banach. It yields boundedness; indeed, the same formula verifies the sharper estimate \(\|Tf\|_\infty\leq\|f\|_\infty\).
The sequential argument is useful here because convergence in the domain controls the output at every transformed point \(x^2\), without requiring any separate analysis of the graph as a subset of a product space.
Worked Example: Differentiation Is Closed but Not Bounded on an Incomplete Domain
Let \(D=C^1([0,1])\), regarded as a subspace of \(C([0,1])\) with the inherited supremum norm, and define \(S:D\to C([0,1])\) by \(Sf=f'\). The graph is closed in \(C([0,1])\times C([0,1])\): suppose \(f_n\to f\) uniformly and \(f_n'\to g\) uniformly, where each \(f_n\in D\). For every \(x\in[0,1]\), the fundamental theorem of calculus gives \[ f_n(x)=f_n(0)+\int_0^x f_n'(t)\,dt. \] Taking limits, using uniform convergence on both sequences, gives \[ f(x)=f(0)+\int_0^x g(t)\,dt. \] Since \(g\) is continuous, this identity implies \(f\in C^1([0,1])\) and \(f'=g\). Thus \((f,g)\) belongs to the graph of \(S\).
Nevertheless, \(S\) is unbounded for the inherited supremum norm. For each positive integer \(n\), let \(f_n(x)=x^n\). Then \(\|f_n\|_\infty=1\), while \(\|Sf_n\|_\infty=\|n x^{n-1}\|_\infty=n\). No finite constant \(C\) can satisfy \(\|Sf\|_\infty\leq C\|f\|_\infty\) for every \(f\in D\).
There is no contradiction with the Closed Graph Theorem: \(D\) is not complete in the inherited norm. For example, \(h_n(x)=\sqrt{x+1/n}\) belongs to \(D\) and converges uniformly to \(\sqrt{x}\), since \[ 0\leq\sqrt{x+1/n}-\sqrt{x}\leq 1/\sqrt{n} \] for \(x\in[0,1]\). The limit \(\sqrt{x}\) is not in \(C^1([0,1])\). Completeness of the domain is essential to the theorem.
What Closedness Does and Does Not Say
The graph condition is stronger than merely knowing that each output \(Tx\) exists. It constrains what happens when inputs and outputs converge together. In the Closed Graph Theorem, that constraint, combined with completeness, supplies the continuity that might otherwise require a direct estimate.
A frequent pitfall is to apply the theorem to a map whose domain is only a subspace of a Banach space. The theorem requires the domain, with its stated norm, to be a Banach space and the operator to be defined on all of that domain. Differentiation in the example is closed as an operator from \(C^1([0,1])\) with the inherited norm, but that domain is incomplete. If a proper domain is equipped with a different, complete norm, the boundedness conclusion concerns that norm, not necessarily the norm inherited from a larger space.
Another important point is that closedness must be checked in the product topology determined by the specified norms. Changing the norm on the domain can change completeness and can change whether the graph is closed. The theorem is therefore a statement about an operator together with its domain and norm, not just its algebraic formula.
Check Your Understanding
Use the graph, projection, and completeness arguments to answer the following questions.
- In the sequential criterion, why must both \(x_n\to x\) and \(Tx_n\to y\) be assumed?
- Why is the graph \(\Gamma(T)\) a Banach space when it is closed in \(X\times Y\) and both \(X\) and \(Y\) are Banach?
- Where does the proof use that \(T\) is defined on all of \(X\)?
- How does boundedness of the inverse projection from \(X\) to \(\Gamma(T)\) yield boundedness of \(T\)?
- Why can differentiation have a closed graph and still be unbounded in the third worked example?