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Let $X$ be a normed linear space over $\mathbb C$ and let $X^*$ denote its dual. For each $x\in X$ we have a linear map $\Phi_x:X^*\to \mathbb C$ which sends $f\in X^*$ to $f(x)$. Recall that the weak* topology on $X^*$ is the coarsest topology such that $\Phi_x$ is continuous for each $x\in X$.

Question. Suppose $\Phi:X^*\to\mathbb C$ is a linear map which is continuous with respect to the weak* topology. Is it necessarily true that $\Phi=\Phi_x$ for some $x\in X$.

I suspect the answer to the above question is 'yes' because the answer in this post seems to suggest so. However, I am stuck.

caffeinemachine
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1 Answers1

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That follows from a more general statement:

Let $X'$ be a subspace of all linear functionals on $X$. Let $\tau$ be the topology for $X$ induced by $X'$. Then the continuous linear functionals of $X$ with respect to $\tau$ is exactly $X'$.

For $f_{0}$ a continuous linear functional with respect to $\tau$, then there is a neighborhood of $0$ in $X$ mapped by $f_{0}$ into $\{|x|<1\}$. Then there are neighborhoods $U_{1},...,U_{n}$ of $0$ in $X$ and $f_{1},...,f_{n}\in X'$ such that $f_{0}(f_{1}^{-1}(U_{1})\cap\cdots\cap f_{n}^{-1}(U_{n}))\subseteq\{|x|<1\}$.

For $x\in\ker(f_{1})\cap\cdots\cap\ker(f_{n})$, we have $|f_{0}(mx)|=m|f_{0}(x)|<1$ for all $m=1,2,...$, this forces $f_{0}(x)$ to be zero, and hence $x\in\ker(f)$.

A classic linear algebra fact states that $f=\displaystyle\sum_{i=1}^{n}c_{i}f_{i}$ for some constants $c_{i}$, and hence $f\in X'$.

Now every $f\in X'$ is certainly continuous with respect to $\tau$ by the very definition of induced topology $\tau$.

Now let $Q(X)=\{Q(x): x\in X\}$, where $\left<x^{\ast},Q(x)\right>=\left<x,x^{\ast}\right>$ for $x^{\ast}\in X'$ and $x\in X$. Then the weak$^{\ast}$ topology for $X^{\ast}$ is exactly the topology induced by $Q(X)$.

user284331
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