If we have a norm $\lVert\cdot\rVert_{norm}$, then we know that the dual norm is defined as $$\lVert x\rVert_{norm}^{*}:=\sup_{\lVert y \rVert_{norm}=1}x^{T}y.$$ Furthermore a generalized Cauchy-Schwartz Inequality holds $$x^{T}y\leq \lVert x\rVert_{norm}^{*} \lVert y\rVert_{norm}.$$ What happens now if instead of a norm we have a seminorm $\lVert\cdot\rVert_{semi}$? Does a dual still exist? Can it be defined in the same way? And does the generalized Cauchy-Schwartz Inequality still hold?
Does a dual seminorm (similar to dual norms) exist for seminorms? And does the generalized Cauchy-Schwartz inequality still hold?
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real-analysis
functional-analysis
norm
normed-spaces
1 Answers
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This fails super hard, even if the semi-norms actually induce a norm topology. Consider for example $\Bbb C^2$ with semi-norms: $$\|(x_1,x_2)\|_1=|x_1|\qquad\|(x_1,x_2)\|_2=|x_2|$$ You have then for example $\|(0,x)\|_1=0$ for all $x$, so if you take the dual element $x_2^*:(x_1,x_2)\mapsto x_2$ then $$\sup_{x\in\Bbb C, \|x\|_1≤1}|x_2^*(x)|=\sup_{(x_1,x_2)\in\Bbb C\, |x_1|<1}|x_2|=\infty$$
Another thing you could ask, in the situation that you have semi-norms $\|\cdot\|_\alpha$ is whether or not $$\sup_{x\in V\,\|x\|_{\alpha}≤1\,\forall\alpha}|x^*(x)|<\infty$$ I believe this should be true.