Let $U$ be a closed of $H$ , then is subset of $H$

Question

How to prove it ?

I was reading following pdf and this question is based on examples in it.

source : https://sites.math.washington.edu/~burke/crs/555/555_notes/hilbert.pdf

Answer 1

Let $V$ be the set in question.

$V$ is closed: if $(v_n)$ is a convergent sequence in $V$, then $v_n=x+y_n$ and $y_n$ is convergent to some $y$ in $U$, since $U$ is closed. Then $v_n\rightarrow x+y\in V$, so $V$ is closed.

$V$ is convex: let $(x+y)$,$(x+z)$ be two elements of $V$. Then for any $\lambda\in[0,1]$ we have $$\lambda(x+y)+(1-\lambda)(x+z)=x+(\lambda y+(1-\lambda) z)\in V$$ where the second term is in $U$ because $U$ is a subspace. So V is convex.