The Big Three Pt. 6 - Closed Graph Theorem with Applications

(Before everything: elementary background in topology and vector spaces, in particular Banach spaces, is assumed.)

A surprising result of Banach spaces

We can define several relations between two norms. Suppose we have a topological vector space $X$ and two norms $‖ \cdot ‖_{1}$ and $‖ \cdot ‖_{2}$ . One says $‖ \cdot ‖_{1}$ is weaker than $‖ \cdot ‖_{2}$ if there is $K > 0$ such that $‖ x ‖_{1} \leq K ‖ x ‖_{2}$ for all $x \in X$ . Two norms are equivalent if each is weaker than the other (trivially this is a equivalence relation). The idea of stronger and weaker norms is related to the idea of the “finer” and “coarser” topologies in the setting of topological spaces.

So what about their limit? Unsurprisingly this can be verified with elementary $ϵ - N$ arguments. Suppose now $‖ x_{n} - x ‖_{1} \to 0$ as $n \to 0$ , we immediately have

‖ x_{n} - x ‖_{2} \leq K ‖ x_{n} - x ‖_{1} < K ε

for some large enough $n$ . Hence $‖ x_{n} - x ‖_{2} \to 0$ as well. But what about the converse? We give a new definition of equivalence relation between norms.

(Definition) Two norms $‖ \cdot ‖_{1}$ and $‖ \cdot ‖_{2}$ of a topological vector space are compatible if given that $‖ x_{n} - x ‖_{1} \to 0$ and $‖ x_{n} - y ‖_{2} \to 0$ as $n \to \infty$ , we have $x = y$ .

By the uniqueness of limit, we see if two norms are equivalent, then they are compatible. And surprisingly, with the help of the closed graph theorem we will discuss in this post, we have

(Theorem 1) If $‖ \cdot ‖_{1}$ and $‖ \cdot ‖_{2}$ are compatible, and both $(X, ‖ \cdot ‖_{1})$ and $(X, ‖ \cdot ‖_{2})$ are Banach, then $‖ \cdot ‖_{1}$ and $‖ \cdot ‖_{2}$ are equivalent.

This result looks natural but not seemingly easy to prove, since one find no way to build a bridge between the limit and a general inequality. But before that, we need to elaborate some terminologies.

Preliminaries

(Definition) For $f : X \to Y$ , the graph of $f$ is defined by
$G (f) = {(x, f (x)) \in X \times Y : x \in X} .$

If both $X$ and $Y$ are topological spaces, and the topology of $X \times Y$ is the usual one, that is, the smallest topology that contains all sets $U \times V$ where $U$ and $V$ are open in $X$ and $Y$ respectively, and if $f : X \to Y$ is continuous, it is natural to expect $G (f)$ to be closed. For example, by taking $f (x) = x$ and $X = Y = R$ , one would expect the diagonal line of the plane to be closed.

(Definition) The topological space $(X, τ)$ is an $F$ -space if $τ$ is induced by a complete invariant metric $d$ . Here invariant means that $d (x + z, y + z) = d (x, y)$ for all $x, y, z \in X$ .

A Banach space is easily to be verified to be a $F$ -space by defining $d (x, y) = ‖ x - y ‖$ .

(Open mapping theorem) See this post

By definition of closed set, we have a practical criterion on whether $G (f)$ is closed.

(Proposition 1) $G (f)$ is closed if and only if, for any sequence $(x_{n})$ such that the limits
$x = lim_{n \to \infty} x_{n} and y = lim_{n \to \infty} f (x_{n})$
exist, we have $y = f (x)$ .

In this case, we say $f$ is closed. For continuous functions, things are trivial.

(Proposition 2) If $X$ and $Y$ are two topological spaces and $Y$ is Hausdorff, and $f : X \to Y$ is continuous, then $G (f)$ is closed.

Proof. Let $G^{c}$ be the complement of $G (f)$ with respect to $X \times Y$ . Fix $(x_{0}, y_{0}) \in G^{c}$ , we see $y_{0} \neq f (x_{0})$ . By the Hausdorff property of $Y$ , there exists some open subsets $U \subset Y$ and $V \subset Y$ such that $y_{0} \in U$ and $f (x_{0}) \in V$ and $U \cap V = \emptyset$ . Since $f$ is continuous, we see $W = f^{- 1} (V)$ is open in $X$ . We obtained a open neighborhood $W \times U$ containing $(x_{0}, y_{0})$ which has empty intersection with $G (f)$ . This is to say, every point of $G^{c}$ has a open neighborhood contained in $G^{c}$ , hence a interior point. Therefore $G^{c}$ is open, which is to say that $G (f)$ is closed. $◻$

closed-graph

REMARKS. For $X \times Y = R \times R$ , we have a simple visualization. For $ε > 0$ , there exists some $δ$ such that $| f (x) - f (x_{0}) | < ε$ whenever $| x - x_{0} | < δ$ . For $y_{0} \neq f (x_{0})$ , pick $ε$ such that $0 < ε < \frac{1}{2} | f (x_{0}) - y_{0} |$ , we have two boxes ( $C D E F$ and $G H J I$ on the picture), namely

B_{1} = {(x, y) : x_{0} - δ < x < x_{0} + δ, f (x_{0}) - ε < y < f (x_{0}) + ε}

and

B_{2} = {(x, y) : x_{0} - δ < x < x_{0} + δ, y_{0} - ε < y < y_{0} + ε} .

In this case, $B_{2}$ will not intersect the graph of $f$ , hence $(x_{0}, y_{0})$ is an interior point of $G^{c}$ .

The Hausdorff property of $Y$ is not removable. To see this, since $X$ has no restriction, it suffices to take a look at $X \times X$ . Let $f$ be the identity map (which is continuous), we see the graph

G (f) = {(x, x) : x \in X}

is the diagonal. Suppose $X$ is not Hausdorff, we reach a contradiction. By definition, there exists some distinct $x$ and $y$ such that all neighborhoods of $x$ contain $y$ . Pick $(x, y) \in G^{c}$ , then all neighborhoods of $(x, y) \in X \times X$ contain $(x, x)$ so $(x, y) \in G^{c}$ is not a interior point of $G^{c}$ , hence $G^{c}$ is not open.

Also, as an immediate consequence, every affine algebraic variety in $C^{n}$ and $R^{n}$ is closed with respect to Euclidean topology. Further, we have the Zariski topology $Z$ by claiming that, if $V$ is an affine algebraic variety, then $V^{c} \in Z$ . It’s worth noting that $Z$ is not Hausdorff (example?) and in fact much coarser than the Euclidean topology although an affine algebraic variety is both closed in the Zariski topology and the Euclidean topology.

The closed graph theorem

After we have proved this theorem, we are able to prove the theorem about compatible norms. We shall assume that both $X$ and $Y$ are $F$ -spaces, since the norm plays no critical role here. This offers a greater variety but shall not be considered as an abuse of abstraction.

(The Closed Graph Theorem) Suppose

(a) $X$ and $Y$ are $F$ -spaces,

(b) $f : X \to Y$ is linear,

(c) $G (f)$ is closed in $X \times Y$ .

Then $f$ is continuous.

In short, the closed graph theorem gives a sufficient condition to claim the continuity of $f$ (keep in mind, linearity does not imply continuity). If $f : X \to Y$ is continuous, then $G (f)$ is closed; if $G (f)$ is closed and $f$ is linear, then $f$ is continuous.

Proof. First of all we should make $X \times Y$ an $F$ -space by assigning addition, scalar multiplication and metric. Addition and scalar multiplication are defined componentwise in the nature of things:

α (x_{1}, y_{1}) + β (x_{2}, y_{2}) = (α x_{1} + β x_{2}, α y_{1} + β y_{2}) .

The metric can be defined without extra effort:

d ((x_{1}, y_{1}), (x_{2}, y_{2})) = d_{X} (x_{1}, x_{2}) + d_{Y} (y_{1}, y_{2}) .

Then it can be verified that $X \times Y$ is a topological space with translate invariant metric. (Potentially the verifications will be added in the future but it’s recommended to do it yourself.)

Since $f$ is linear, the graph $G (f)$ is a subspace of $X \times Y$ . Next we quote an elementary result in point-set topology, a subset of a complete metric space is closed if and only if it’s complete, by the translate-invariance of $d$ , we see $G (f)$ is an $F$ -space as well. Let $p_{1} : X \times Y \to X$ and $p_{2} : X \times Y \to Y$ be the natural projections respectively (for example, $p_{1} (x, y) = x$ ). Our proof is done by verifying the properties of $p_{1}$ and $p_{2}$ on $G (f)$ .

For simplicity one can simply define $p_{1}$ on $G (f)$ instead of the whole space $X \times Y$ , but we make it a global projection on purpose to emphasize the difference between global properties and local properties. One can also write $p_{1} |_{G (f)}$ to dodge confusion.

Claim 1. The projection $p_{1}$ (with restriction on $G (f)$ ) defines an isomorphism between the image of $p_{1}$ on $G (f)$ and $X$ .

For $x \in X$ , we see $p_{1} (x, f (x)) = x$ (surjectivity). If $p_{1} (x, f (x)) = 0$ , we see $x = 0$ and therefore $(x, f (x)) = (0, 0)$ , hence the restriction of $p_{1}$ on $G$ has trivial kernel (injectivity). Further, it’s trivial that $p_{1}$ is linear.

Claim 2. $p_{1}$ is continuous on $G (f)$ .

For every sequence $(x_{n})$ such that $lim_{n \to \infty} x_{n} = x$ , we have $lim_{n \to \infty} f (x_{n}) = f (x)$ since $G (f)$ is closed, and therefore $lim_{n \to \infty} p_{1} (x_{n}, f (x_{n})) = x$ . Meanwhile $p_{1} (x, f (x)) = x$ . The continuity of $p_{1}$ is proved.

Claim 3. $p_{1}$ is a homeomorphism with restriction on $G (f)$ .

We already know that $G (f)$ is an $F$ -space, so is $X$ . For $p_{1}$ we have $p_{1} (G (f)) = X$ is of the second category (since it’s an $F$ -space and $p_{1}$ is one-to-one), and $p_{1}$ is continuous and linear on $G (f)$ . By the open mapping theorem, $p_{1}$ is an open mapping on $G (f)$ , hence is a homeomorphism.

Claim 4. $p_{2}$ is continuous.

This follows the same way as the proof of claim 2 but much easier since there is no need to care about $f$ .

Now things are immediate once one realises that $f = p_{2} \circ p_{1} |_{G (f)}^{- 1}$ , which implies that $f$ is continuous. $◻$

Applications

Before we go for theorem 1 at the beginning, we mention an application on Hilbert spaces.

Let $T$ be a bounded operator on the Hilbert space $L_{2} ([0, 1])$ so that if $ϕ \in L_{2} ([0, 1])$ is a continuous function so is $T ϕ$ . Then the restriction of $T$ to $C ([0, 1])$ is a bounded operator of $C ([0, 1])$ .

For details please check this.

Now we go for the identification of norms. Define

\begin{aligned} f : (X, ‖ \cdot ‖_{1}) & \to (X, ‖ \cdot ‖_{2}) \\ x & \mapsto x \end{aligned}

i.e. the identity map between two Banach spaces (hence $F$ -spaces). Then $f$ is linear. We need to prove that $G (f)$ is closed. For the convergent sequence $(x_{n})$

lim_{n \to \infty} ‖ x_{n} - x ‖_{1} = 0,

we have

lim_{n \to \infty} ‖ f (x_{n}) - x ‖_{2} = lim_{n \to \infty} ‖ x_{n} - x ‖_{2} = lim_{n \to \infty} ‖ f (x_{n}) - f (x) ‖_{2} = 0.

Hence $G (f)$ is closed. Therefore according to the closed graph theorem, $f$ is continuous, hence bounded, we have some $K$ such that

‖ x ‖_{2} = ‖ f (x) ‖_{1} \leq K ‖ x ‖_{1} .

By defining

\begin{aligned} g : (X, ‖ \cdot ‖_{2}) & \to (X, ‖ \cdot ‖_{1}) \\ x & \mapsto x \end{aligned}

we see $g$ is continuous as well, hence we have some $K^{'}$ such that

‖ x ‖_{1} = ‖ g (x) ‖_{2} \leq K^{'} ‖ x ‖_{2}

Hence two norms are weaker than each other.

The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it’s time to make a list of the series. It’s been around half a year.

The Big Three Pt. 1 - Baire Category Theorem Explained
The Big Three Pt. 2 - The Banach-Steinhaus Theorem
The Big Three Pt. 3 - The Open Mapping Theorem (Banach Space)
The Big Three Pt. 4 - The Open Mapping Theorem (F-Space)
The Big Three Pt. 5 - The Hahn-Banach Theorem (Dominated Extension)
The Big Three Pt. 6 - Closed Graph Theorem with Applications

References

Walter Rudin, Functional Analysis
Peter Lax, Functional Analysis
Jesús Gil de Lamadrid, Some Simple Applications of the Closed Graph Theorem

updated at 2025-05-27

# The big three