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Let $X$ be a normed space and let $W\subset X^*$ be a subspace which separates the points in $X$. Let $\psi \in X^*$ such that $\ker \psi $ is $W$-weakly closed. Show that $\psi \in W$.

Any ideas?

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Took me a minute. First a pure linear algebra lemma:

Lemma. Suppose $X$ is a vector space over $K$ and $\lambda_1,\dots,\lambda_n,\psi$ are linear functionals on $X$. If $\bigcap_{j=1}^n \ker(\lambda_j)\subset \ker(\psi)$ then $\psi$ is a linear combination of $\lambda_1,\dots,\lambda_n$.

Proof: Define $\Lambda:X\to K^n$ by $\Lambda x=(\lambda_1x,\dots,\lambda_nx)$. Let $Y=\Lambda(X)$. Now, $\psi x$ depends only on $\lambda_1x,\dots,\lambda_nx$, meaning that if $\lambda_jx=\lambda_jx'$ for every $j$ then $\psi x=\psi x'$. So there exists $T:Y\to K$ so that $\psi x=T(\Lambda x)$ for every $x\in X$. It is clear that $T$ is linear on $Y$. So $T$ extends to a linear map from $K^n$ to $K$. We know what those are; there exist $\alpha_j\in K$ so $T(\Lambda x)=\sum\alpha_j\lambda_jx$. QED

Now for your problem. If $\psi=0$ then $\psi\in W$; assume $\psi\ne0$. Choose $x\in X$ with $\psi x\ne0$.

The complement of $\ker \psi$ is $W$-open; by definition of the $W$ topology this says that there exist $\lambda_1,\dots,\lambda_n\in W$ and $\epsilon>0$ so that if $|\lambda_jx-\lambda_jx'|<\epsilon$ for every $j$ then $\psi x'\ne0$.

Now suppose that $y\in\bigcap_{j=1}^n\ker(\lambda_j)$. Then $\psi(x+\alpha y)\ne0$ for every scalar $\alpha$ by the above with $x'=x+\alpha y$. This is impossible unless $\psi(y)=0$.

So $\bigcap_{j=1}^n\ker(\lambda_j)\subset\ker(\psi)$, hence $\psi$ is a linear combination of the $\lambda_j$.

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