Does uniform boundedness principle hold for other types of operator norm?

Question

Let $X$ be a Banach space and $(x_\alpha^*)_\alpha$ be a net in $X^*$, where $X^*$ is the continuous dual of $X$. For each $\alpha$, let $\|x_\alpha^*\|_{op}$ be the operator norm of $x_\alpha^*$, i.e., $$\|x_\alpha^*\|_{op} = \sup_{x_\alpha\in X, \|x_\alpha\|\leq 1}|x_\alpha^*(x_\alpha)|.$$

To show that $$\sup_\alpha \|x_\alpha^*\|_{op}<\infty,$$ by the uniform boundedness principle, it suffices to show that $$\sup_\alpha|x_\alpha^*(x)| < \infty \quad \text{for all} \quad x\in X.$$

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I am interested whether the same arguments above hold for $x_\alpha^*$ having two components.

More precisely, for each $\alpha$, let $x_\alpha^* = (x_{\alpha 1}^*, x_{\alpha 2}^*):X\times X\to \mathbb{R}$ be defined by $$x_\alpha^*(x_1,x_2) = x_{\alpha 1}^*(x_1) + x_{\alpha 2}^*(x_2).$$ The norm on $x_\alpha^*$ is defined by $$\|x_\alpha^*\|_1 = \|x_{\alpha 1}^*\|_{op} + \|x_{\alpha 2}^*\|_{op}.$$ Now, suppose that $(x_\alpha^*)$ is a net in $(X\times X)^*$. I want to show that $$\sup_\alpha\| x_\alpha^* \|_1 < \infty.$$ Does the uniform boundedness principle hold in this case, i.e., is it sufficient to show that $$\sup_\alpha |x_\alpha^*(x)|<\infty \quad \text{for all} \quad x = (x_1,x_2)\in X\times X?$$ To summarize, my question is,

if $$\sup_\alpha |x_{\alpha 1}^*(x_1) + x_{\alpha 2}^*(x_2)| < \infty \quad \text{for all}\quad (x_1,x_2)\in X\times X,$$ does it imply that $$\sup_\alpha (\|x_{\alpha 1}^*\|_{op} + \|x_{\alpha 2}^*\|_{op})<\infty?$$

Answer 1

Just apply the hypothesis to $(x_1,0)$ and $(0, x_2)$. You can then apply UBP separately to the two components.

Answer 2

Yes.

Your norm $\lVert-\rVert_1$ is the operator norm when you equip $X\times X$ with the norm $\lVert(x,y)\rVert=\max\{\lVert x\rVert,\lVert y\rVert\}$.