Linear mapping between Banach space and its dual

Question

This problem has been giving me some troubles. Does anyone have any ideas on how to go about proving this?

Let X be a Banach space and let $$\Phi: X \rightarrow X^*$$ be a linear mapping such that for each $x\in X$ we have $\Phi(x)(x)=0.$

Then:

For all $x,y\in X $we have $\Phi(x)(y)= -\Phi(y)(x) $ and $\Phi $ is bounded.

I suspect to use Uniform Boundedness Principle which says:

Suppose that F is a collection of continuous linear operators from X to Y. If

$$\sup\nolimits_{T \in F} \|T(x)\|_Y < \infty, $$

for all $x\in X$, then

$$\sup\nolimits_{T \in F,\|x\|=1} \|T(x)\|_Y=\sup\nolimits_{T \in F} \|T\|_{B(X,Y)} < \infty.$$

Answer 1

The first part has been answered by rae306.

For the second part apply Closed Graph Theorem. Suppose $x_n \to x$ and $\Phi(x_n) \to f$. Then $\Phi (x_n) (y)=-\Phi (y) (x_n)$. Taking limits we get $f(y)=-\Phi (y) (x)$. [ Note that $\Phi (y)$ is continuous for fixed $y$]. Applying the first part again we get $f(y)=\Phi (x) (y)$. This is true for each $y$ so $f=\Phi (x)$ as required.