Brezis Functional Analysis Exercise 2.4

Let $\alpha: E \times F \to \mathbb{R}$ ($E,F$ banach spaces) be a linear functional with the following property:

For fixed $x \in E$, $\alpha$ is a continuous map.

For fixed $y \in F$, $\alpha$ is a continuous map.

Prove that there exists $C \geq 0$ such that for all $x,y \in E,F$ respectively, we have:$$ |a(x,y)| \leq C ||x||y|| $$

At first this question seemed to be a relatively straightforward application of the schwarz inequality, however, Brezis gives the following hint:

And the corollary is:

I guess my question here is mostly about corollary 2.5: How does one define an inner product between two elements of different spaces? I.e, is it on me to define the inner product $<f,x>$? Could I just use $\alpha$ and then evaluate it at $x$ to define a suitable inner product? I thought of most of this while typing the preceding sentences, so apologies in advance for not attempting it.

1 Answer

Let be $T: S_E \to F^*$ the linear map such that $\langle T(x),y\rangle = \alpha(x,y)$, $\forall y \in F$ and $\forall x \in E$. Since that from hypothesis for all fixed $y \in F$, there are $C_y > 0$ such that

$$|\langle T(x), y\rangle | = |\alpha(x,y)| \leq C_y\|x\| = C_y$$

follows from Corollary 2.5 there are $C > 0$ that not depend of $x \in E$ and $y \in F$ such that $$ |\alpha(x,y)| = |\langle T(x),y \rangle | \leq C\|y\|, \forall x \in S_E$$

and in particular, if $ x \neq 0 \in E$

$$ \left|\alpha\left(\frac{x}{|x|},y\right)\right| \leq C|y|$$ implying, by linearity of $\alpha$ , that

$$ |\alpha(x,y)| \leq C|y||x|.$$

Observetion: Here, $S_E$ is unitary sphere of $E$.