Let $(V,\left \langle .,. \right \rangle)$ be a space of Hilbert and $F\in V'$. Then $U=\{v\in V: F(v)=0\}$ is a closed subspace of $V$.

Question

Let $(V,\left \langle .,. \right \rangle)$ be a space of Hilbert and $F\in V'$. Then $U=\{v\in V: F(v)=0\}$ is a closed subspace of $V$.

I have thought of doing the following: Let's take $\{x_n\}_n\subset U$ a sequence and prove that $x_n\to x$ in $U$. Note that $F(x_n)=0$ for all $n\in \mathbb{N}$ so $F(x_n)\to 0$. But I don't know how to prove that $\{x_n\}_n$ converges and that this limit is an element of $U$, any idea? Thank you!

Answer 1

An alternative way is to see that $U=F^{-1}(\{0\})$ and $\{0\}$ is a closed set and the pre-image of a closed set of a continuity map is still closed.

Answer 2

You don't have to prove that $(x_n)$ converges. To prove that $U$ is closed you have to take sequence $x_n$ in $U$ which converges to some $u$ and then prove that $u \in U$. This is obvious from continuity of $F$. Convergence is given to you.

Answer 3

Whenever $T : X\to Y$ is a bounded operator between normed spaces, then $\ker T$ is closed. Proof: Let $x_n\in\ker T$ such that $x_n\to x$. Then, by continuity of $T$, we have $Tx = \lim_nTx_n = 0$. So, $x\in\ker T$.

A bounded linear functional is just a special case of a bounded operator.