Whenever $f=T(x)$ then $f(x) \geq 0$. Prove $T$ is bounded by closed graph theorem.

Question

Let X be a Banach space and suppose $T:X \rightarrow X^*$ is linear and has the property that whenever $f=T(x)$ then $f(x)\geq 0$. Prove T is bounded.

The book gives a hint: using closed graph theorem.

I stucked at proving $|T(x_n)-T(x)|=\sup_{\|y\|\leq1}|f_{x_n}(y)-f_x(y)|\rightarrow0$.

Any suggestion is appriciated.

Answer 1

Let $x_n\to0$ and suppose $T(x_n)\to F$, which is not zero. As such there must be some $y\in X$ with $F(y)>0$. Now:

$$0 ≤ T(\lambda y-x_n)\,[\lambda y - x_n] = \lambda^2 T(y)\,[y] - \lambda T(x_n)\,[y] + T(\lambda y- x_n)\,[x_n],$$ in particular the limit of this expresiion, which is equal to $\lambda^2 T(y)\,[y]- \lambda F(y)$, must be $≥0$ for all $\lambda$. For small $\lambda$ this is impossible, as $F(y)>0$.

This contradiction implies that the operator $T$ is closed.