Considering $E$ a normed vector space. 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 \} $
How can we prove that $F(x)$ is also:
$F(x) = \Big\{ f \in E^*; \dfrac{1}{2}\|y\|^2 - \dfrac{1}{2}\|x\|^2 \geq \langle f, y-x \rangle \ \forall y \in E \Big\} $
I cannot figure it out. Please help me. Thanks.
On
This problem requires a prerequisite that
Lemma: $F(x)$ can be written as $$F(x)=\{f\in E^{*}:\|f\|\leq\|x\|\ \text{and}\ <f,x>=\|x\|^{2}\}.$$
I will assume the lemma to prove this problem and prove the lemma.
Proof of the Problem:
Define $$N(x):=\Big\{f\in E^{*}:\dfrac{1}{2}\|y\|^{2}-\dfrac{1}{2}\|x\|^{2}\geq <f, y-x>,\forall y\in E\Big\}.$$ To prove $F(x)\subset N(x)$, we need to recall that, for any $a,b\in\mathbb{R}$, we have $$2ab\leq a^{2}+b^{2},$$ and the Cauchy-Schwarz Inequality in the inner product space that for any vectors $u,v$ in an inner product space, we have $$|<u,v>|^{2}\leq<u,u>\cdot<v,v>,$$ such that we have $$|<u,v>|\leq\|u\|\ \|v\|.$$
Now, let $f\in F(x)$, then we have \begin{align*} <f,y-x>&=<f,y>-<f,x>\\ &=<f,y>-\|x\|^{2},\ \text{since}\ f\in F(x)\\ &\leq|<f,y>|-\|x\|^{2}\\ &\leq\|f\|\ \|y\|-\|x\|^{2},\ \text{by}\ \text{Cauchy-Schwarz Inequality}\\ &=\|x\|\ \|y\|-\|x\|^{2},\ \text{since}\ f\in F(x)\\ &\leq\dfrac{\|x\|^{2}}{2}+\dfrac{\|y\|^{2}}{2}-\|x\|^{2},\ \text{by the firstly introduced inequality}\\ &=\dfrac{\|y\|^{2}}{2}-\dfrac{\|x\|^{2}}{2}. \end{align*}
Therefore, $f\in N(x)$, and thus $F(x)\subset N(x).$
Conversely, let $f\in N(x)$, since the inequality described in $N(x)$ holds for all $y\in E$, it must hold for $y:=xt$ for all $t\in\mathbb{R}$. Thus, we have $$<f,tx-x>\leq\dfrac{1}{2}\|xt\|^{2}-\dfrac{1}{2}\|x\|^{2},$$ such that $$(t-1)<f,x>\leq\dfrac{1}{2}(t^{2}-1)\|x\|^{2},\ \text{for all}\ t\in\mathbb{R}.$$
If $t>1$, divide $(t-1)$ on both side and then we have $$<f,x>\leq\dfrac{1}{2}(t+1)\|x\|^{2},$$ taking $t\rightarrow 1^{+}$ yields us $$<f,x>\leq \|x\|^{2}.$$
On the other hand, if $t<1$, dividing $(t-1)$ on both side yields us $$<f,x>\geq\dfrac{1}{2}(t+1)\|2\|^{2},$$ taking $t\rightarrow 1^{-}$ yields us $$<f,x>\geq\|x\|^{2}.$$
Therefore, $<f,x>=\|x\|^{2}.$
Now, plugging in this concluded equality into the inequality described in $N(x)$ yields us $$<f,y>\leq\dfrac{1}{2}\|y\|^{2}+\dfrac{1}{2}\|x\|^{2},\ \text{for all}\ y\in E.$$
Thus, if we choose $y\in E$ such that $\|y\|\leq\delta>0$, where $\delta$ is arbitrarily fixed, then we have $$<f,y>\leq \dfrac{1}{2}\delta^{2}+\dfrac{1}{2}\|x\|^{2},\ \text{for all such}\ y,$$ and thus we have $$\delta\|f\|:=\sup_{\substack{y\in E \\ \|y\|\leq 1}}<f, \delta y>=\sup_{\substack{z\in E \\ \|z\|\leq\delta}}<f,z>\leq\dfrac{1}{2}\delta^{2}+\dfrac{1}{2}\|x\|^{2}.$$
Thus, if we choose $\delta=\|x\|$, we have $$\|f\|\leq \|x\|.$$
It follows from the Lemma that $f\in F(x)$, and thus $N(x)\subset F(x)$.
Proof of Lemma:
Set $$G(x):=\{f\in E^{*}:\|f\|\leq\|x\|\ \text{and}\ <f,x>=\|x\|^{2}\}.$$
Then, $F(x)\subset G(x)$ for sure.
Conversely, let $f\in G(x)$, if $x=0$, then $\|f\|\leq 0\implies \|f\|=0$, so let's suppose $x\neq 0.$ Then, $$<f,x>=\|x\|^{2}\implies<f,\dfrac{1}{\|x\|}x>=\|x\|,$$ but $$\Big\|\dfrac{x}{\|x\|}\Big\|=1,$$ and thus by definition $$\|f\|=\sup_{\substack{\|y\|\leq 1 \\ y\in E}}|<f,y>|\geq \Big|<f,\dfrac{1}{\|x\|}x>\Big|=\|x\|.$$
Thus, $\|f\|\geq \|x\|$, and hence $\|f\|=\|x\|.$
Thus, $f\in F(x)$, and therefore $F(x)\supset G(x).$
The second definition of $F$ coincides with the definition of the subdifferential of the convex function $x\mapsto \frac12\|x\|^2$. It is a standard exercise to prove that this is equivalent to your first definition.
Let me prove that $\|f\|=\|x\|$ and $\langle f,x\rangle =\|x\|^2$ implies that $f$ belongs to the second set. Take $y\in X$. Then $$ \langle f,y-x\rangle = \langle f,y\rangle - \|x\|^2 \le \frac 12\|f\|^2 + \frac12\|y\|^2 - \|x\|^2 = \frac12\|y\|^2 -\frac 12\|x\|^2. $$ Let now $f$ be in the second set. Take $v\in X$, $\lambda\in\mathbb R$, set $y=x+\lambda v$. Then $$ 0\ge \langle f,y-x\rangle - \frac12\|y\|^2 +\frac12\|x\|^2 = \lambda \langle f,v\rangle - \frac12\|x+\lambda v\|^2 +\frac12\|x\|^2 \\ = \lambda \langle f,v\rangle - \frac12 (\|x+\lambda v\|-\|x\|) (\|x+\lambda v\|+\|x\|) $$ which implies $$ \lambda \langle f,v\rangle \le \frac12 (\|x+\lambda v\|-\|x\|) (\|x+\lambda v\|+\|x\|) \le \frac12\lambda \|v\|(\|x+\lambda v\|+\|x\|) $$ Dividing by $|\lambda|$ and let $\lambda\to 0$ gives $$ \langle f,v\rangle \le \|v\|\cdot \|x\|. $$ Setting $v=x$ in the above calculation yields $$ \lambda \langle f,x\rangle \le \frac12 ((\lambda+1)^2-1)\|x\|^2 = \frac12 (\lambda^2+2\lambda)\|x\|^2, $$ dividing by $\lambda<0$ and let $\lambda\nearrow0$ $$ \langle f,x\rangle \ge \|x\|^2, $$ which proves $\|f\|=\|x\|$ and $\langle f,x\rangle=\|x\|^2$.