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Let $\langle\;\cdot\;,\;\cdot\;\rangle$ be a duality pairing between a Banach space $X$ and a normed space $Y$ with $$\left|\langle x,y\rangle\right|\le\left\|x\right\|_X\left\|y\right\|_Y\;\;\;\text{for all }(x,y)\in X\times Y\tag1.$$

Are we able to show that $$\left\|x\right\|_X=\sup_{\left\|y\right\|_Y\le1}\left|\langle x,y\rangle\right|\tag2$$ for all $x\in X$?

Let $x\in X\setminus\{0\}$. By $(1)$, $$\varphi:=\langle x,\;\cdot\;\rangle\in Y'\tag3$$ and $$c:=\sup_{\left\|y\right\|_Y\le1}\left|\langle x,y\rangle\right|=\left\|\varphi\right\|_{Y'}\le\left\|x\right\|_X\tag4.$$

On the other hand, since $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing, there is a $y\in Y$ with $$\varphi(y)\ne0;\tag5$$ but at this point I'm stuck.


Motivation: If $E$ is a normed space, it is an immediate consequence of the Hahn-Banch theorem that, for all $x\in E$, there is a $\varphi\in E'$ with $\varphi(x)=\left\|x\right\|_E$ and $\left\|\varphi\right\|_{E'}\le1$. This yields that $$\langle x,\varphi\rangle:=\varphi(x)\;\;\;\text{for }(x,\varphi)\in E\times E'$$ is a duality pairing between $E$ and $E'$ satisfying $$\left\|x\right\|_E=\sup_{\|\varphi\|_{E'}\le1}|\langle x,\varphi\rangle|\;\;\;\text{for all }x\in E.\tag6$$

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    $\begingroup$ Suppose $\Phi: X \times Y \to \mathbb{R}$ is a duality pairing satisfying $(1)$ and $(2)$. Isn't $\frac12 \Phi$ a duality pairing that satisfies $(1)$ but not $(2)$? $\endgroup$ Dec 24 '20 at 15:34
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As stated in the comments, $\frac{1}{2}\Phi$ is a duality pairing which satisfies $(1)$ but not $(2)$. Although, the norm is bounded above by the duality pairing in the sense that $$||x||_{X}=\sup_{||y||\leq 2}\frac{1}{2}|\Phi(x,y)|$$ There are some duality pairings $\Phi :X\times Y\to \mathbb{R}$ for which $$\tag{*}\bigl|\Phi(x,y)\bigr|\leq ||x||_{X}\cdot ||y||_{Y}$$ But there is no constant $C>0$ such that $$\tag{**} ||x||_{X}\leq C\sup_{||y||\leq 1}\bigl|\Phi(x,y)\bigr|$$ For example, if $X=Y=\ell_2$ and $\Phi:X\times Y\to \mathbb{R}$ is given by $$\Phi(x,y)=\sum_{n=1}^{\infty}\frac{1}{n}x(n) y(n)$$ $x=(x(n))_{n=1}^{\infty},\,y=(y(n))_{n=1}^{\infty}\in \ell_2$. Then, by the Cauchy-Schwarz inequality \begin{align} \bigl|\Phi(x,y)\bigr|&=\biggl|\sum_{n=1}^{\infty}\frac{1}{n}x(n)y(n)\biggr|\leq \sum_{n=1}^{\infty}\biggl|\frac{1}{n}x(n)y(n)\biggr|\\ &\leq \biggl(\sum_{n=1}^{\infty}\frac{1}{n^2}(x(n))^2\biggr)^{1/2}\cdot \biggl(\sum_{n=1}^{\infty}(y(n))^2\biggr)^{1/2}\\ &\leq \biggl(\sum_{n=1}^{\infty}(y(n))^2\biggr)^{1/2}\cdot \biggl(\sum_{n=1}^{\infty}(x(n))^2\biggr)^{1/2}=||x||_2\cdot ||y||_2 \end{align} Hence, $(1)$ is satisfied. Now, consider the sequence $me_m$ where $e_m$ is the standard basis of $\ell_2$. I.e $e_m(n)=1$ if $n=m$ and $e_m(n)=0$ for every $n\neq m$. Then, for every $||y||_{2}\leq 1$ $$\Phi(me_m,y)=\frac{1}{m}m e_m(m)\cdot y(m)=y(m)$$ Hence, $\bigl|\Phi(me_m,y)\bigr|=|y(m)|\leq ||y||_{2}=1$ which means that $\sup_{||y||_2\leq 1}\bigl|\Phi(me_m,y)\bigr|\leq 1$. If there exists $C>0$ such that $(**)$ holds then we end up with \begin{align} m&=||me_m||_{2}\leq C\sup_{||y||_{2}\leq 1}\bigl|\Phi(me_m,y)\bigr|\leq C \end{align} for all $m$, which is obviously false.

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