Prove that if the dual space is strictly convex, then the duality map contains a single point

Let $$E$$ be be a vector space over $$\mathbb{R}$$, and denoted by $$E^{*}$$ the dual space of it, i.e. the space all of continuous linear functional on $$E$$. Then, the duality map $$F$$ is defined for every $$x\in E$$ by $$F(x):=\{f\in E^{*}:\|f\|=\|x\|\ \text{and}\ \langle f,x \rangle =\|x\|^{2}\}.$$

I am asked to show that

If $$E^{*}$$ is strictly convex, then $$F(x)$$ contains a single point.

This is the Q2 of exercise section 1.1 of Chapter 1 of H. Brezis' Book "Functional Analysis, Sobolev Spaces, and Partial Differential Equations".

I understand that this problem can be shown in one line, by just stating, as suggested in the book, that

In a strictly convex normed space, any nonempty convex set that is contained in a sphere is reduced to a single point.

However, before I read the solution in the book, I tried the following elementary way which may not need the above fact, but I was stuck.

Here is what I have done:

Suppose $$F(x)$$ contains at least two points. Let $$f,g\in F(x)$$ be such that $$f\neq g$$, and let $$0<\lambda<1$$ be an arbitrary real number.

Then, since $$F(x)$$ is convex, we have $$\lambda f+(1-\lambda)g\in F(x).$$

Thus, we have $$\|f\|=\|g\|=\|x\|=\|\lambda f+(1-\lambda)g\|,$$ so that if $$\|f\|=\|g\|=1$$, then we have $$\|\lambda f+(1-\lambda)g\|=1,$$ which contradicts the hypothesis that $$E^{*}$$ is strictly convex.

Thus, $$F(x)$$ contains at most $$1$$ point, but $$F(x)$$ is not empty, so $$F(x)$$ contains only one point.

This solution seems only able to conclude that $$F(x)$$ contains a single point for those $$x$$ on the unit ball, but not for all $$x\in E$$.

Is there a way to extend this proof to all $$x\in E$$?

I am really new to functional analysis.

Yes, my question is really a dumb one. I forgot the condition in the definition of strict convexity.

Recall the definition of strict convexity:

If $$f,g\in E^{*}$$ such that $$f\neq g$$ and $$\|f\|=\|g\|=1$$, then for $$0<\lambda<1$$, we have $$\|\lambda f+(1-\lambda)g\|< 1.$$

Thus, this proof actually begins with choosing $$f,g\in E^{*}$$ such that $$f\neq g$$ and $$\|f\|=\|g\|=1$$, not assuming $$\|f\|=\|g\|=1$$ in the middle. Then, everything follows smoothly:

Suppose $$F(x)$$ contains at least two points. Let $$f,g\in F(x)$$ be such that $$f\neq g$$ and $$\|f\|=\|g\|=1$$, and let $$0<\lambda<1$$ be an arbitrary real number.

Then, since $$F(x)$$ is convex, we have $$\lambda f+(1-\lambda)g\in F(x).$$

Thus, we have $$\|f\|=\|g\|=\|x\|=\|\lambda f+(1-\lambda)g\|=1,$$ which contradicts the hypothesis that $$E^{*}$$ is strictly convex.

Thus, $$F(x)$$ contains at most $$1$$ point, but $$F(x)$$ is not empty, so $$F(x)$$ contains only one point.

You can't just assume that $$\left\lVert{f}\right\lVert=\lVert{g}\lVert=1$$ because $$\lVert{x}\lVert$$ could be different from $$1$$. What you have to do is divide every thing by $$\lVert{x}\lVert$$

$$\lVert{f}\lVert=\lVert{g}\lVert=\lVert{x}\lVert=\lVert{\lambda f + (1-\lambda)g}\lVert$$ $$\left\lVert{\frac{f}{\lVert{x}\lVert}}\right\lVert=\left\lVert{\frac{g}{\lVert{x}\lVert}}\right\lVert=1=\left\lVert{\lambda \frac{f}{\lVert{x}\lVert} + (1-\lambda)\frac{g}{\lVert{x}\lVert}}\right\lVert<1$$

• Yes you are right. Commented Nov 28, 2022 at 21:17