Sum of closed spaces is not closed

Question

I recently encountered the theorem that the sum of a closed (linear) subspace with a finite dimensional subspace is closed subspace of the Banach space in which it is contained. However, this came with the caveat that the statement does not hold for two arbitrary closed subspaces.

So, here's what I'm looking for:

Find a Banach space $X$ and closed subspaces $M,N$ such that $$ M+N=\{m+n\mid m\in M, n\in N\} $$ Is not closed in $X$.

Any references, hints, or answers are appreciated!

Answer 1

Let $X$ be a Hilbert space with orthonormal basis $\{e_n\}_{n\in\mathbb{N}}$. Let $x_n=e_{2n}$, and let $y_n=e_{2n}+\frac{e_{2n+1}}{n+1}$. Take $M$ to be the closed span of the $x_n$ and $N$ to be the closed span of the $y_n$. Note that $M+N$ contains $e_n$ for all $n$, so the closure of $M+N$ is all of $X$.

However, I claim that $M+N$ does not contain the vector $z=\sum \frac{e_{2n+1}}{n+1}$ and hence is not all of $X$. Indeed, if you could write $z=x+y$ for $x\in M$ and $y\in N$, it is clear that $y$ would have to be $\sum y_n$, since the only way to get a nonzero inner product with $e_{2n+1}$ when building an element of $M$ or $N$ is to use $y_n$. Since the sum $\sum y_n$ does not converge, there are no such $x$ and $y$.

Answer 2

Let $X=c_0$ be the space of sequences $\left(α_j\right)$ that converge to zero, equipped with the sup norm. Consider the subsets $$ Y=\left\{\left(α_j\right)\middle|α_{2 j-1}=0, j=1,2, …\right\} \text{ and } Z=\left\{\left(α_j\right)\middle|α_{2 j}=j^2 α_{2 j-1}, j=1,2, …\right\} $$ Since $Y$ and $Z$ are the intersection of kernels of some bounded linear functionals, $Y$ and $Z$ are closed subspaces of $X$.

We will show that the element $x=\left(1,0, \frac14, 0, \frac19, 0, …\right)$ lies in the closure of $Y+Z$ but not in $Y+Z$.

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To prove $x∉Y+Z$, suppose $x=y+z,y∈Y,z∈Z$, then $z_{2j-1}=x_{2j-1}-y_{2j-1}=\frac1{j^2},z_{2j}=j^2z_{2j-1}=1$ contradicting $z_j→0$.

To prove $x∈\overline{Y+Z}$, let $x_n=\sum_{j=1}^n\frac1{j^2}e_{2j-1}$ then $x_n→x$ since $${‖x-x_n‖}=\left\|\sum_{j=n+1}^∞\frac1{j^2}e_{2j-1}\right\|=\frac1{(n+1)^2}→0$$ For all $j$, $\tfrac{1}{j^{2}} e_{2j-1}=(\frac1{j^2}e_{2j-1}+e_{2j})-e_{2j}∈Y+Z$, so $x∈\overline{Y+Z}$.

Answer 3

Here is a general construction of such a pair of closed subspaces for any infinite-dimensional Hilbert space. This construction can be found in Section 1.15 of Introduction to Hilbert Space and the Theory of Spectral Multiplicity (Second Edition) by Halmos. Note that the solution provided by Eric Wofsey resembles a particular case of this construction.

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space. Let $(e_{n})_{n\in\mathbb{N}}$ and $(f_{n})_{n\in\mathbb{N}}$ be two orthonormal sequences in $\mathcal{H}$ such that $(e_{n}, f_{m})_{\mathcal{H}} = 0$ for all $m,n\in\mathbb{N}$. Let $(\alpha_{n})_{n\in\mathbb{N}}$ and $(\beta_{n})_{n\in\mathbb{N}}$ be sequences in $\mathbb{K}$ such that $\beta_{n}\neq 0$ for all $n\in\mathbb{N}$, $|\alpha_{n}|^{2} + |\beta_{n}|^{2} = 1$ for all $n\in\mathbb{N}$ and $\sum_{n\in\mathbb{N}} |\beta_{n}|^{2} < \infty$. (For an explicit example, take $\beta_{n} = \tfrac{1}{n}$ and $\alpha_{n} = (1 - \tfrac{1}{n^{2}})^{\tfrac{1}{2}}$ for each $n\in\mathbb{N}$.) Define $g_{n} := \alpha_{n} e_{n} + \beta_{n} f_{n}$ for each $n\in\mathbb{N}$. Using the assumption $|\alpha_{n}|^{2} + |\beta_{n}|^{2} = 1$ for each $n\in\mathbb{N}$, it is straightforward to check that $(g_{n})_{n\in\mathbb{N}}$ is an orthonormal sequence in $\mathcal{H}$.

Define $M := \overline{{\rm span}}\{e_{n} : n\in\mathbb{N}\}$ and $N := \overline{{\rm span}}\{g_{n} : n\in\mathbb{N}\}$. Then $M$ and $N$ are closed subspaces of $\mathcal{H}$. Define $u := \sum_{n\in\mathbb{N}} \beta_{n} f_{n}$, where the sum converges (unconditionally) because $\sum_{n\in\mathbb{N}} |\beta_{n}|^{2} < \infty$. We claim $u\in \overline{M+N}$ and $u\not\in M + N$, from which follows that $M + N$ is not closed in $\mathcal{H}$.

We first show $u\in \overline{M+N}$. For each $k\in\mathbb{N}$ we have $\beta_{k}\neq 0$ and hence $f_{k} = - \alpha_{k} \beta_{k}^{-1} e_{k} + \beta_{k}^{-1} g_{k} \in M+N$, so it follows that $\sum_{k=1}^{n} \beta_{k} f_{k} \in M+N$ for each $n\in\mathbb{N}$ and therefore $u\in \overline{M+N}$.

We now show $u\not\in M+N$. Suppose for a contradiction this is not the case. Write $u = x + y$ with $x\in M$ and $y\in N$. Let $n\in\mathbb{N}$. Observe that $f_{n}\in M^{\perp}$, $(g_{n}, f_{n})_{\mathcal{H}} = \beta_{n}$ and $(g_{m}, f_{n})_{\mathcal{H}} = 0$ for all $m\in\mathbb{N}\setminus\{n\}$. Hence we have \begin{align*} \beta_{n} = (u, f_{n})_{\mathcal{H}} = (x + y, f_{n})_{\mathcal{H}} = (y, f_{n})_{\mathcal{H}} &= (\sum_{k\in\mathbb{N}} (y, g_{k})_{\mathcal{H}} g_{k}, f_{n} )_{\mathcal{H}} \\ &= (y, g_{n})_{\mathcal{H}} (g_{n}, f_{n})_{\mathcal{H}} = \beta_{n} (y, g_{n})_{\mathcal{H}} . \end{align*} As $\beta_{n} \neq 0$ for each $n\in\mathbb{N}$, this implies $(y, g_{n})_{\mathcal{H}} = 1$ for all $n\in\mathbb{N}$. But this contradicts the Bessel inequality. Therefore $u\not\in M+N$.