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