Htykuut

Exercise :

Let $M, N$ be closed subspaces of the Hilbert space $H$ while it is also assumed that $M bot N$. Show that the set :
$$M + N = {x+y : x in M, y in N}$$
is also a closed subspace of $H$.

Attempt - Thoughts :

First of all, since $M$ and $N$ are closed, this means that if $(x_n)$ is a sequence of elements belonging in $X$ and $(y_n)$ is a sequence of elements belonging in $N$, then :
$$lim x_n = x, quad x in M$$
$$lim y_n = y, quad y in N$$
Now, since it is $M bot N$, then it should also be $langle x, y rangle = 0$.

Now, defining the Hilbert space norm to be $|cdot | = sqrt{langle cdot, cdot rangle}$, it is :

$$|x+y|^2 = |x|^2 + 2langle x,yrangle + |y||^2$$

But since $x bot y$, then : $ |x+y|^2 = |x|^2 + |y|^2 Leftrightarrow |x+y| = sqrt{|x|^2 + |y|^2}$.

In terms of the definition of a closed set, I would need to prove that for $(x_n) in M$ and $(y_n) in N$, then $(x_n + y_n) in M + N$ and $lim (x_n + y_n) in M + N$.

I can't really see though how I could proceed to any conclusions (or if there's a simpler way etc), so I would really appreciate any hints or elaborations

Suppose ${x_n} subset M$,${y_n} subset N$ and $x_n+y_nto z$. Then (by orthogonality) $|(x_n-x_m) -(y_n-y_m)|^{2}=|x_n-x_m|^{2}+|y_n-y_m|^{2} to 0$ so both ${x_n}$ and ${y_n}$ are Cauchy. If these sequences converge to $x$ and $y$ respectively the $x in M, yin N$ and $z=x+y in M+N$.

$M=(M^perp)^perp$ and $N=(N^perp)^perp$. Then $M+N=(M^perpcap N^perp)^perp$. As the perpendicular complement of a subspace is always closed, then
$M+N$ is closed.

You can as well run through the definitions.
Assume ${x_n+y_n}_{n=1}^{infty}subseteq M+N$ a Cauchy sequence then using $Mperp N$ gives
$$langle x_n+y_n-(x_m+y_m),x_n+y_n-(x_m+y_m)rangle=langle x_n-x_m,x_n-x_mrangle +langle y_n-y_m,y_n-y_mranglerightarrow 0$$
as $n,mrightarrowinfty$ from which You can conclude that ${x_n}$ and ${y_n}$ are Cauchy sequences and thus have limits $xin M$ and $yin N$ respectively since $M$ and $N$ are closed.
Since for the considered sequences one has $langle x_n-x,x_n-xranglerightarrow 0,nrightarrow infty$ and $langle y_n-y,y_n-yranglerightarrow 0,nrightarrowinfty$, You can conclude
$$langle x_n+y_n-(x+y),x_n+y_n-(x+y)rangle=langle x_n-x,x_n-xrangle +langle y_n-y,y_n-yranglerightarrow 0$$
as $nrightarrowinfty$ and thus $M+N$ is closed.

Worked around the following answer :

Let $(z_n)$ be a convergent sequence in $M + N$. Then we can write $z_n = underbrace{x_n}_{in M} +underbrace{y_n}_{in N}$.

But $M perp N $ we have $| z_n |^2 = |x_n |^2 +|y_n|^2$ by the Pythagoreian Theorem, since $langle x_n,y_nrangle = 0$ .

Now, the sequence $(z_n)$ is a Cauchy sequence, which means that :
$$ |x_n -x_m|^2 +|y_n -y_m|^2 = | z_n -z_m|^2 rightarrow 0$$

since $x_n to x$ and $ y_nto y$ for some $xin M $ , $y in N $ since they are both closed spaces . It then follows that $$lim z_n = z = x+y in N + M $$
thus the space $N+M$ is a closed subspace of $H$.