If $Z_n = X_n + Y_n$ for $X_n\in M$ and $Y_n\in N$ then $(X_n)$ and $(Y_n)$ converge

Question

Let $H$ be a Hilbert space (infinite dim) with $M,N\subset H$ being closed subspaces satisfying $N\subset M^\perp$.

I'm trying to show that $M+N$ is closed.

If $(Z_n)_{n=1}^\infty \subset M+N$ is a sequence then $Z_n = X_n+Y_n$ for some sequences $(X_n)_{n=1}^\infty$ and $(Y_n)_{n=1}^\infty$ in $M$ and $N$.

I can show $M+N$ is closed if I know that $X_n\to X$ and $Y_n\to Y$ for some $X\in M$ and $Y\in N$, so my question is: why do these two sequences converge?

Answer 1

From the orthogonality of $Y_n-Y_m$ and $X_n-X_m$ we get $\|(X_n+Y_n)-(X_m+Y_m)\|^{2} =\|X_n-X_m\|^{2}+\|Y_n-Y_m\|^{2}$. Hence $\|X_n-X_m\|^{2} \leq \|(X_n+Y_n)-(X_m+Y_m)\|^{2}\to 0$. So the Cauchy sequence $X_n$ converges to some $X$. Since $X_n+Y_n$ also converges we see that $Y_n$ converges too. Can you finish?

Answer 2

We suppose $Z_n = X_n + Y_n \to Z$

Note that the projections

$$\pi_1: H \to M$$ $$\pi_2: H \to M^\perp$$

are continuous (by the closed graph theorem).

Thus $X_n = \pi_1(Z_n) \to \pi_1(Z) \in M$ and similarly $Y_n \to \pi_2(Z) \in M^\perp$.

But $Y_n \in N$ and $N$ is closed, so $\pi_2(Z) \in N$.

Answer 3

Choose a sequence $(z_{n})$ in $M+N$ and suppose that $z_{n}\rightarrow z\in\mathcal{H}$ for some $z$. Write $z_{n}=x_{n}+y_{n}$, where $x_{n}\in M$ and $y_{n}\in N$. (Note that $x_{n}$ and $y_{n}$ are unique). For any $m<n$, we have \begin{eqnarray*} & & \left\Vert z_{n}-z_{m}\right\Vert ^{2}\\ & = & \left\Vert (x_{n}-x_{m})+(y_{n}-y_{m})\right\Vert ^{2}\\ & = & \left\Vert x_{n}-x_{m}\right\Vert ^{2}+\left\Vert y_{n}-y_{m}\right\Vert ^{2}. \end{eqnarray*} Since $(z_{n})$ is a Cauchy sequence, from the above identity, it is clear that $(x_{n})$ and $(y_{n})$ are also Cauchy sequences. Therefore $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$ for some $x,y\in\mathcal{H}$. Notice that $M$ and $N$ are closed, so $x\in M$ and $y\in N$. It follows that $z=x+y\in M+N$.