If $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing satisfying the Cauchy-Schwarz inequality, is $\|x\|=\sup_{\|y\|\le1}|\langle x,y\rangle|$

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 $$\left\|x\right\|_E=\sup_{\|\varphi\|_{E'}\le1}|\langle x,\varphi\rangle|\;\;\;\text{for all }x\in E.\tag6$$

• 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 $(2)$? Dec 24 '20 at 15:34

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.