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Let $V$ be a real Banach space and $T:V\to V^{\ast }$ be linear with $T_v(v)\ge 0$ for every $v$. If $v_n\to 0$ and $T_{v_n}\to f$ in $V^{\ast }$, positivity of $T_{v_n+tw}(v_n+tw)$ and passage to the limit givefor every real $t$ and every $w$. Taking small $t$ of either sign forces $f(w)=0$. Thus the graph of $T$ is closed, and the closed graph theorem makes $T$ continuous.

Let $X\subseteq {\mathbb{R}}^n$ be Borel. If $L^q(X)\subseteq L^p(X)$ for $1\le p<q\le \infty$, the closed graph theorem makes the inclusion continuous. Applying its norm bound to indicators of $X\cap B(0,R)$ shows that their measures are uniformly bounded, hence $\mu (X)<\infty$.

Regard the set-theoretic inclusionas a linear map between Banach spaces. Its graph is closed. Indeed, if $f_n\to f$ in $L^q$ and $f_n\to g$ in $L^p$, part (i) gives an almost-everywhere convergent subsequence for the first convergence and then a further subsequence converging almost everywhere for the second; hence $f=g$ almost everywhere. The closed graph theorem therefore gives

Let $E_R=X\cap B(0,R)$. It has finite measure, and $\mu (E_R)\uparrow \mu (X)$. Applying the bound to $1_{E_R}$ gives, for finite $q$,so $\mu (E_R{)}^{1/p-1/q}\le C$. For $q=\infty$, it gives $\mu (E_R{)}^{1/p}\le C$. In either case the measures $\mu (E_R)$ are uniformly bounded. Continuity from below now yieldsThis proves the converse summarized by Lq inclusion implies finite measure on a Euclidean Borel subset.

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.

DefineThe hypothesis makes $T$ everywhere defined. The dual $V^{\ast }$ is Banach even when $V$ is incomplete. If $f_j\to f$ in $V^{\ast }$ and $T{f}_j\to a$ in ${\ell }^1$, then coordinatewise convergence gives $a_n=f(x_n)$, so the graph of $T$ is closed. The closed graph theorem therefore givesTaking the supremum over the unit ball proves the required finite bound.

The open mapping theorem states that every surjective bounded linear map $S:X\to Y$ between Banach spaces maps open sets to open sets. Equivalently, there is $c>0$ such that

The closed graph theorem states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its graphis closed in $X\times Y$.

Assume the open mapping theorem and suppose $\Gamma (T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projectionis a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma (T)\to Y$ is bounded, andThus $T$ is bounded and hence continuous. This is the closed graph theorem from the bounded inverse theorem argument.

Write $T_v=T(v)$. We prove that $T$ has a closed graph. SupposePut $u_n=v_n-v$ and $g=f-T_v$. Then $u_n\to 0$ and $T_{u_n}\to g$ in $V^{\ast }$.

Fix $w\in V$ and $t\in \mathbb{R}$. Positivity and linearity giveBecause $(T_{u_n})$ converges in the continuous dual space, its norms are bounded, soAlso $T_{u_n}(w)\to g(w)$, while $T_w(u_n)\to 0$ because the fixed functional $T_w$ is continuous. Passing to the limit yieldsfor every real $t$. If $g(w)\ne 0$, a sufficiently small $t$ of the opposite sign makes the right-hand side negative. Hence $g(w)=0$ for every $w$, so $g=0$ and $f=T_v$.

The graph is therefore closed. The closed graph theorem now proves that $T:V\to V^{\ast }$ is continuous. This is precisely continuity of a positive linear map into a dual space.

If $X$ is a reflexive Banach space, its closed unit ball is weakly compact. A bounded linear map is weak-to-weak continuous, so $T(B_X)$ is weakly compact in $Y$. Compact subsets of a Hausdorff space are closed, hence $T(B_X)$ is weakly closed and therefore norm closed.

Now suppose the two norms on $V$ have the same continuous dual as a set. Each $X_j^{\ast }$ is a Banach space in its dual norm. The identityhas closed graph: if $f_n\to f$ in the first dual norm and $f_n\to g$ in the second, evaluating at each $x\in V$ gives $f(x)=g(x)$. The closed graph theorem makes $I$ bounded, and the same argument for $I^{-1}$ makes the dual norms equivalent. Thus there are $c,C>0$ withThe dual formulafrom the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent.

The bounded inverse theorem states that a bounded bijective linear map between Banach spaces has a bounded inverse.

The closed graph theorem states that if $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is continuous if and only if its graphis closed in the Banach space $X\times Y$.

If $T$ is continuous and $(x_n,T{x}_n)\to (x,y)$, then $T{x}_n\to Tx$, so uniqueness of limits gives $y=Tx$. Hence the graph is closed.

Conversely, suppose $\Gamma (T)$ is closed. It is then a Banach space. The coordinate projectionis bounded and bijective. By the bounded inverse theorem, $P_X^{-1}:x\mapsto (x,Tx)$ is bounded. Composing it with the bounded second-coordinate projection givesso $T$ is continuous.

Consider the inclusion mapIts graph is closed. Indeed, suppose $x_k\to x$ in $X$ and $J{x}_k\to y$ in $L^1$. By hypothesis, some subsequence of $(x_k)$ converges almost everywhere to $x$. From its $L^1$ convergence, that subsequence has a further subsequence converging almost everywhere to $y$. Uniqueness of almost-everywhere limits gives $x=y$ as $L^1$ classes.

Both spaces are Banach, so the closed graph theorem makes $J$ bounded. Therefore some $C>0$ satisfies