I know that if $X$ is a Banach space, then, the direct sum of two closed subspace $X_1$ and $X_2$ is not necessarily closed. But what if $X$ is Hilbert?

I assume there is something to do with the orthonormal basis, since this is something you can ask for a Hilbert space but not a Banach space.

Moreover, is the projection to $X_1$ bounded? Namely, $P$ is defined on $X_1\oplus X_2$: $$P(x)=x\quad \text{on $X_1$} \qquad P(x)=0\quad \text{on $X_2$}$$ is $P$ bounded?

  • $\begingroup$ There is a counter-example given by Jim Belk here: math.stackexchange.com/questions/82300/…. $\endgroup$ – Davide Giraudo Apr 22 '12 at 19:55
  • $\begingroup$ The second question seems to have nothing to do with the first question: the answer is: the orthogonal projection onto a closed subspace is always bounded (it has norm $1$). $\endgroup$ – t.b. Apr 22 '12 at 21:39
  • $\begingroup$ @t.b. I'm a little confused. Are all projections orthogonal? $\endgroup$ – henryforever14 Apr 22 '12 at 22:01
  • $\begingroup$ @t.b never mind. projections are not necessarily orthogonal. So, that if the projection is not orthogonal, is it bounded? $\endgroup$ – henryforever14 Apr 22 '12 at 22:10
  • $\begingroup$ Sorry: I completely misread the second part of your question. I suggest that you accept Robert Israel's answer that addresses both questions and gives a much simpler example. $\endgroup$ – t.b. Apr 22 '12 at 23:03

In the sequence space $\ell^2$, let $X_1$ be the subspace of sequences $x = (x_0,x_1,x_2, \ldots)$ such that $x_{2n} = 0$ for all natural numbers $n$, and $X_2$ the subspace of sequences $x$ such that $x_{2n+1} = n x_{2n}$ for all $n$. It is easy to see that $X_1 \cap X_2 = \{0\}$, and that $X_1 + X_2$ is dense (e.g. it contains all sequences with finite support). However, $X_1 + X_2$ is not all of $\ell^2$, e.g. it can't contain the sequence $x_n = 1/(n+1)$: if this was $u + v$ with $u \in X_1$ and $v \in X_2$, we'd need $v_{2n} = x_{2n} = 1/(2n+1)$ and $v_{2n+1} = n/(2n+1)$, but then $\sum_n |v_n|^2 = \infty$.

When $X_1 \oplus X_2$ is not closed, the projection $P$ of $X_1 \oplus X_2$ onto $X_1$ must be unbounded: otherwise it would extend to a bounded linear operator $Q$ from $\overline{X_1 \oplus X_2}$ onto $X_1$; since $(I - P)x \in X_2$ for $x \in X_1 \oplus X_2$, we'd also have $(I-Q) x \in X_2$ for $x \in \overline{X_1 \oplus X_2}$. But then $x = Qx + (I-Q)x \in X_1 \oplus X_2$ for $x \in \overline{X_1 \oplus X_2}$, contradiction.

  • $\begingroup$ Thanks! It is a very nice example. $\endgroup$ – henryforever14 Apr 23 '12 at 0:36

Let $X = H_{0}^1(0,1) \mathbin{\oplus_2} L^2(0,1)$ be the Hilbert space direct sum of the Sobolev space $H_{0}^1(0,1)$ (the completion of $C_{c}^\infty(0,1)$ with respect to the norm $\|f\|_{H^1} = (\|f\|_{L^2}^2 +\|f'\|_{L^2}^2)^{1/2}$) and the usual Hilbert space $L^2(0,1)$. Then $X$ is a Hilbert space with respect to the norm $\|(f,g)\| = (\|f\|_{H^1}^2 + \|g\|_{L^2}^2)^{1/2}$.

Let $i: H^1(0,1) \to L^2(0,1)$ be the natural inclusion. The graph $\Gamma = \{ (f,f)\,:\,f \in H_{0}^1\} \subset X$ of the function $i: H_{0}^1 \to L^2$ is closed because $i$ is bounded: $\|i(f)\|_{L^2} = \|f\|_{L^2} \leq \|f\|_{H^{1}}$. The subspace $U = H_{0}^{1}(0,1) \mathbin{\oplus_2} \{0\}$ is also closed in $X$. I claim that $\Gamma + U$ is not closed in $X$.

The image $i(H_{0}^1) \subset L^2$ is dense because $H_{0}^1(0,1)$ contains $C_{c}^\infty(0,1)$, and $i$ is not onto because functions in $H_{0}^1$ have absolutely continuous representatives. Therefore there is $g \in L^2 \smallsetminus H_{0}^1$ and there are $f_n \in H_{0}^1$ such that $\|g - f_n\|_{L^2} \to 0$. Now note that $(0,f_n) = (f_n,f_n) + (-f_n,0) \in \Gamma + U$, so $(0,g)$ is in the closure of $\Gamma + U$ but not in $\Gamma + U$ itself.

This is a special case of my answer here adapted to the Hilbert space setting.


Exercise 1.20, p. 40 in Rudin FA also presents such $X_1,X_2$ in a Hilbert space.

BTW, whenever $X_1,X_2$ are subspaces of a TVS, $X_1$ closed and $X_2$ finite-dimensional, $X_1+X_2$ is closed. This is Theorem 1.42 (p. 30) in Rudin FA.

Rudin: Functional Analysis, 2nd edition, 1991. (Its TVSs are Hausdorff.) https://archive.org/details/RudinW.FunctionalAnalysis2e1991


A general class of counterexamples is given by Exercises II.3.9 and II.3.10 of [Conway]; see references below.

Exercise II.3.9. Let $A \in \mathscr B(\mathscr H)$ and $\mathscr N = \operatorname{graph}(A) \subseteq \mathscr H \oplus \mathscr H$, that is, $\mathscr N = \{h \oplus Ah \, : \, h\in\mathscr H\}$. Because $A$ is continuous and linear, $\mathscr N \leqslant \mathscr H \oplus \mathscr H$. Let $\mathscr M = \mathscr H \oplus (0) \leqslant \mathscr H \oplus \mathscr H$. Prove the following statements: (a) $\mathscr M \cap \mathscr N = (0)$ if and only if $\ker(A) = (0)$. (b) $\mathscr M + \mathscr N$ is dense in $\mathscr H \oplus \mathscr H$ if and only if $\operatorname{ran}(A)$ is dense in $\mathscr H$. (c) $\mathscr M + \mathscr N = \mathscr H \oplus \mathscr H$ if and only if $A$ is surjective.

(Here the notation $V \leqslant W$ means that $V$ is a closed linear subspace of $W$.)

The proofs of the statements in the preceding exercise are straightforward. This exercise is followed, suggestively, by the following.

Exercise II.3.10. Find two closed linear subspaces $\mathscr M,\mathscr N$ of an infinite-dimensional Hilbert space $\mathscr H$ such that $\mathscr M \cap \mathscr N = (0)$ and $\mathscr M + \mathscr N$ is dense in $\mathscr H$, but $\mathscr M + \mathscr N \neq \mathscr H$.

Of course, the solution is to give an example of a Hilbert space $\mathscr H$ and an operator $A \in \mathscr B(\mathscr H)$ with $\ker(A) = (0)$ such that $\operatorname{ran}(A)$ is dense in $\mathscr H$, but $\operatorname{ran}(A) \neq \mathscr H$. Then choose $\mathscr M$ and $\mathscr N$ as in Exercise II.3.9, and the result follows.

A clear example of such an operator is the operator $A : \ell^2 \to \ell^2$ given by $e_n \mapsto \frac{1}{n + 1}e_n$, where $\{e_0,e_1,\ldots\} \subseteq \ell^2$ denotes the standard orthonormal basis. This choice of $\mathscr H$ and $A$ gives rise to a counterexample similar to the one given by Robert Israel.

Other, more complicated examples of this form exist as well, as t.b. shows us. (While his answer to the present question gets a little carried away with specifics, the more general construction is mentioned in his answer to another question.)


[Conway]: John B. Conway, A Course in Functional Analysis (1985), Springer Graduate Texts in Mathematics 96.


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