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The problem is

Let $V$ be a Banach space over the real field, and let $T: V \to V^\ast, v \to T_v$ be a linear map between these two Banach spaces that satisfies $T_v(v)\ge 0$ for all $v\in V$. Prove that T is continuous.

I have attempted this question for a while with the guidance of a similar question that have been previously asked here. However, the one I'm trying to prove is more general, so I don't have much progress.

I noticed that if we write $T_v(w)$ as $(v,w)$ then it looks similar to an inner product. We have positivity from the condition $T_v(v)\ge0$ (but not positive definite). We already have (bi)linearity, so it looks similar. This makes me want to have something like Cauchy-Schwarz inequality. Combining with the closed graph theorem, the desired result could perhaps be proved?

S. Chitratta
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1 Answers1

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You don't really need Cauchy-Schwarz, but the Closed Graph Theorem is the way to go. Suppose that $v_n\to v$ and $T_{v_n}\to T$. For any $w\in V$, $$\tag1 0\leq T_{v_n-w}(v_n-w)=T_{v_n}v_n+T_ww-T_{v_n}w-T_wv_n $$ Since convergent sequences are bounded, there exists $c>0$ with $\|v_n\|\leq c$ for all $n$. Then $$ |T_{v_n}v_n-Tv|\leq|(T_{v_n}-T)v_n|+|T(v_n-v)|\leq c\,\|T_{v_n}-T\|+\|T\|\,\|v_n-v\|\to0. $$ This shows that taking limit in $(1)$ we get $$\tag2 0\leq Tv+T_ww-Tw-T_wv=T(v-w)-T_w(v-w). $$ Hence $$\tag3 T_w(v-w)\leq T(v-w),\qquad\qquad w\in V. $$ As $V$ is a vector space, $v-w$ rans over all of $V$ as $w$ rans over all of $V$. So we can rewrite $(3)$ as $$\tag4 T_{v-z}(z)\leq Tz,\qquad\qquad z\in V. $$ Using the linearity of $T$ this we again be rewritten as $$\tag5 (T_v-T)z\leq T_zz,\qquad\qquad z\in V. $$ Replacing $w$ with $w/n$ for $n\in\mathbb Z$, $$\tag6 \pm n(T_v-T)\leq T_ww,\qquad\qquad n\in\mathbb Z, w\in V $$ (note that $T_{w/n}(w/n)=(1/n^2)T_ww$, so the inequality does not change even when $n<0$). The only possibility for $(6)$ to hold is that $T_v-T=0$, which shows that $T_v=T$. Now the Closed Graph Theorem implies that $T$ is bounded.

Martin Argerami
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