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: $$ \bigoplus_{n = 1}^{\infty} \mathscr{H}_n = \left \{\sum_{n = 1}^\infty x_n : x_n \in \mathscr{H}_n, \sum_{n = 1}^\infty\|x_n\|^2 < \infty\right \} $$ 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}^\infty\left |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)+ 2\left(\sum_{n = 1}^\infty|x_n|^2\right)^{1/2} \left(\sum_{n = 1}^{\infty} |y_n|^2\right)^{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?


2 Answers 2


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 $$\sum_{m=1}^k \|x_{m} - x_{1,m}\|^2 = \lim_{n\rightarrow \infty} \sum_{m=1}^k \|x_{n,m} - x_{1,m}\|^2 \le \limsup_{n \rightarrow \infty} \|y_n-y_1\| \le M^2.$$ 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 $$\sum_{m=1}^k \|x_{m} - x_{n',m}\|^2 = \lim_{n\rightarrow \infty} \sum_{m=1}^k \|x_{n,m} - x_{n',m}\|^2 \le \limsup_{n \rightarrow \infty} \|y_n-y_{n'}\|^2 \le \varepsilon^2$$ 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 $\bigoplus_{n = 1}^{\infty} \mathscr{H}_n$.

  • $\begingroup$ Thanks for the answer. I am not really understanding the part where you bounded the sum. What was the reason for doing so?, $\endgroup$ Dec 1, 2018 at 19:06
  • $\begingroup$ Ah I see, you were showing that the sum of squares of the limit sequence $\sum |x_k|^2$ was finite. But couldn't you just say that $y_n$ is cauchy and thus norm bounded? $\endgroup$ Dec 1, 2018 at 21:04
  • $\begingroup$ I want to show that $\sum_{k=1}^\infty \|x_m\| < \infty$ and that is archived by (first) considering a finite sum in order to use the properties of limes (linearity). I.e. we have to argue that we can interchange the limes and the summation. (You can also use Fatou's lemma here for the counting measure on $\mathbb{N}$.) I have extended my answer! $\endgroup$
    – p4sch
    Dec 1, 2018 at 21:23
  • $\begingroup$ I used the monotone convergence theorem. I think that works as well. Thanks. $\endgroup$ Dec 1, 2018 at 21:23
  • $\begingroup$ Yes, indeed! You can also use the monotone convergence theorem. $\endgroup$
    – p4sch
    Dec 1, 2018 at 21:29

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

$$\|x_{m,k} - x_{m,j}\|^2 \le \sum_{m=1}^\infty \|x_{m,k} - x_{m,j}\|^2 = \|y_k - y_j\|^2 \xrightarrow{k,j\to\infty} 0$$ 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 $x_{m,k} \xrightarrow{k\to\infty} x_m$.

Let $\varepsilon > 0$ and pick $N \in \mathbb{N}$ such that $$k,j \ge N \implies \sum_{m=1}^\infty \|x_{m,k} - x_{m,j}\|^2 = \|y_k - y_j\|^2 < \frac\varepsilon2$$

In particular, assuming $k,j \ge N$ for any $K \in \mathbb{N}$ we have $$\sum_{m=1}^K\|x_{m,k} - x_{m,j}\|^2 < \frac\varepsilon2$$

Letting $k \to \infty$ implies $$\sum_{m=1}^K\|x_{m,0} - x_{m,j}\|^2 \le \frac\varepsilon2$$ and since $K$ was arbitrary, it follows $$\sum_{m=1}^\infty\|x_{m,0} - x_{m,j}\|^2 \le \frac\varepsilon2\tag{$*$}$$

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

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 $$j \ge N \implies \|y_0 - y_{m,j}\| \le \frac\varepsilon2 < \varepsilon$$ which means $y_{j} \xrightarrow{j\to\infty} y_0$.


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