# 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 $C$ such that for all $x,y \in E,F$ respectively, we have: $$|a(x,y)| \leq ||x||y||$$

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.

• It's not an inner product - the notation $<f,x>$ just means $f(x)$. – David C. Ullrich Dec 1 '17 at 1:40
• Wow do I feel silly. I had been reading it the whole time as an inner product. I thought it was funny especially because this was the chapter on banach spaces and the notion of inner product had not even been established. Made the book unreadable. Thanks! – rubikscube09 Dec 1 '17 at 1:47
• I don't have that book, but I wouldn't be surprised if the notation was explained when it was first introduced. Look at the first few pages of the chapter... – David C. Ullrich Dec 1 '17 at 1:52
• So the notation section says the symbol $<.,.>$ represents the "scalar product in the duality of $E*,E$", which I am not entirely sure how to interpret. – rubikscube09 Dec 2 '17 at 1:53
• The terminology "scalar product" there is unfortunate. The interpretation he has in mind is just $<f,x>=f(x)$. Honest... – David C. Ullrich Dec 2 '17 at 14:21

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 $E$.

• So here the constant Cx varies with our choice of x (i.e c varies with changes in the functional). Thus, it does not seem to me that Cx bounds the set of values of the functional for fixed y (and varying functionals). Would you be able to explain what I'm misunderstanding here. – rubikscube09 Dec 2 '17 at 2:57
• Sorry, I corrected the arguments. – A.D. Dec 2 '17 at 12:41