Ytukyg

Given a sequence ${mathscr{H}_n}_{n=1}^{infty}$ of closed, orthogonal subspaces of a Hilbert Space $mathscr{H}$, we define the infinite direct sum to be:
$$
bigoplus_{n = 1}^{infty} mathscr{H}_n = left {sum_{n = 1}^infty x_n : x_n in mathscr{H}_n, sum_{n = 1}^infty|x_n|^2 < inftyright }
$$
The question asks me to prove that this is a closed subspace of $mathscr{H}$.

The right hand side condition makes sense to me as for orthogonal $x_n$ we have $|sum x_n|^2 = sum |x_n|^2$. For a sum of two elements in the space, we see that:
begin{align*}
sum_{n = 1}^inftyleft |x_n+ y_n right |^2 &leq sum_{n = 1}^infty (|x_n| + |y_n|)^2 \ &= sum_{n = 1}^infty (|x_n|^2 + 2|x_ny_n| + |y_n|^2) \ &leq sum_{n = 1}^infty (|x_n|^2 + |y_n|^2)+ 2left(sum_{n = 1}^infty|x_n|^2right)^{1/2} left(sum_{n = 1}^{infty} |y_n|^2right)^{1/2} \
&<infty
end{align*}
Thus, a sum of two elements is also a member of the set. the set is also clearly closed under scalar multiplication. How would I prove it is closed? Exactly why can we take sequences and show they converge in the set?

Let $(y_n)_{ n in mathbb{N}}$ be a Cauchy-sequence in $bigoplus_{n = 1}^{infty} mathscr{H}_n$. Write $y_n = sum_{m=1}^infty x_{m,n}$ with $x_{m,n} in mathscr{H}_m$ and note that by using the orthogonality we obtain
$$sum_{m=1}^infty |x_{n,m} - x_{n',m}|^2 = |y_n -y_{n'}|^2.$$
Thus $(x_{n,m})_n$ is also a Cauchy-sequence and since $mathcal{H}_n$ is complete (as a closed set of a complete space), we get that $x_{n,m} rightarrow x_m in mathscr{H}_m$. Next, we show that the sum $sum_{m=1}^infty x_m$ is convergent. (Here we need to argue that we can interchange the limes and the infinite summation.) For this, note that $|y_n - y_1|$ is bounded, say by $M$ and thus
$$sum_{m=1}^k |x_{m} - x_{1,m}|^2 = lim_{nrightarrow infty} sum_{m=1}^k |x_{n,m} - x_{1,m}|^2 le limsup_{n rightarrow infty} |y_n-y_1| le M^2.$$
Hence the last series (on the left hand side) is convergent, because it is bounded. By the $Delta$-inequality we also conclude that $$sum_{m=1}^infty |x_m|^2 <infty.$$ Since $mathscr{H}$ is complete and $(x_m)_m$ are orthogonal, we get that $y = sum_{m=1}^infty x_m$ is convergent in $mathscr{H}$ and by definition we also have $y in bigoplus_{n = 1}^{infty} mathscr{H}_n$. We can take $N in mathbb{N}$ so large that $|y_n -y_{n'}| < varepsilon$ for all $n,n' ge N$. Thus
$$sum_{m=1}^k |x_{m} - x_{n',m}|^2 = lim_{nrightarrow infty} sum_{m=1}^k |x_{n,m} - x_{n',m}|^2 le limsup_{n rightarrow infty} |y_n-y_{n'}|^2 le varepsilon^2$$
for all $n' ge N$. Letting $k rightarrow infty$ shows that
$$|y-y_{n'}|^2 = sum_{m=1}^infty |x_{m} - x_{n',m}|^2 varepsilon^2$$
for all $n' ge N$. Hence $(y_n)_n$ is convegent in $bigoplus_{n = 1}^{infty} mathscr{H}_n$.

As in the answer by @p4sch, let $(y_n)_n$ be a Cauchy sequence in $bigoplus_{n = 1}^{infty} mathscr{H}_n$ with $y_n = sum_{m=1}^infty x_{m,n}$ and $x_{m,n} in mathscr{H}_m$.

For every $m in mathbb{N}$ we have

$$|x_{m,k} - x_{m,j}|^2 le sum_{m=1}^infty |x_{m,k} - x_{m,j}|^2 = |y_k - y_j|^2 xrightarrow{k,jtoinfty} 0$$
so $(x_{m,k})_k$ is Cauchy in $mathscr{H}_m$. Since $mathscr{H}_m$ is complete, there exists $x_{m,0} in mathscr{H}_m$ such that $x_{m,k} xrightarrow{ktoinfty} x_m$.

Let $varepsilon > 0$ and pick $N in mathbb{N}$ such that $$k,j ge N implies sum_{m=1}^infty |x_{m,k} - x_{m,j}|^2 = |y_k - y_j|^2 < fracvarepsilon2$$

In particular, assuming $k,j ge N$ for any $K in mathbb{N}$ we have $$sum_{m=1}^K|x_{m,k} - x_{m,j}|^2 < fracvarepsilon2$$

Letting $k to infty$ implies
$$sum_{m=1}^K|x_{m,0} - x_{m,j}|^2 le fracvarepsilon2$$
and since $K$ was arbitrary, it follows
$$sum_{m=1}^infty|x_{m,0} - x_{m,j}|^2 le fracvarepsilon2tag{$*$}$$

Now we have
$$sum_{m=1}^infty |x_{m,0}|^2 le sum_{m=1}^infty (|x_{m,0}-x_{m,j}| + |x_{m,j}|)^2 le 2left(sum_{m=1}^infty |x_{m,0}-x_{m,j}|^2 + sum_{m=1}^infty|x_{m,j}|^2right) < +infty$$

Hence for $r,s in mathbb{N}$ we have $$left|sum_{m=r}^s x_{m,0}right|^2 = sum_{m=r}^s|x_{m,0}|^2 le sum_{m=r}^infty |x_{m,0}|^2
xrightarrow{r,s to infty} 0$$
so by completeness $y_0 := sum_{m=1}^infty x_{m,0}$ converges in $mathscr{H}$ and by $(*)$ we have $y_0 in bigoplus_{n = 1}^{infty} mathscr{H}_n$.

$(*)$ also means $$j ge N implies |y_0 - y_{m,j}| le fracvarepsilon2 < varepsilon$$ which means $y_{j} xrightarrow{jtoinfty} y_0$.