# Is the distance of two affine subspaces in Hilbert space always attainable?

I have two affine subspaces $$C$$ and $$D$$ in Hilbert space $$\mathcal{H}$$. Define the distance between the two subspaces as $$d(C, D) := \inf_{x\in C, y\in D} d(x, y)$$, where $$d(x, y)$$ is the metric induced by the inner product defined in the Hilbert space. My question is whether this infimum is always attainable. I don't known how to prove either case.

Edit: I've removed the diagonal part since it is not related to this question.

• Do you assume $C$ and $D$ are closed? If not, then surely the answer is no. Aug 6 '19 at 21:34
• Also, what do you mean by "the diagonal of $\mathcal{H}$"? Aug 6 '19 at 21:35
• Eric Wofsey's comment is certainly non-trivial, as evidenced by this question: math.stackexchange.com/questions/1057526/…. Aug 6 '19 at 21:43
• Usually subspace means sub-vector space and we say closed subspace for sub-Hilbert spaces (note a closed subspace is defined by an orthonormal basis but there is no canonical way to define a non-closed infinite dimensional sub vector space, you need to think to it abstractly) Aug 6 '19 at 21:48

This is not true, even for closed affine subspaces. For instance, let $$\mathcal{H}$$ have an orthonormal basis $$(e_n)_{n\in\mathbb{N}}$$. Let $$f_n=e_{2n}+2^ne_{2n+1}$$ and let $$F$$ be the closed span of all the $$f_n$$. Let $$C$$ be the closed span of the $$e_{2n+1}$$, and let $$D$$ be the affine subspace $$v+F$$ where $$v=\sum 2^{-n}e_{2n}$$. Then for any $$x\in\mathcal{H}$$, $$d(x,C)=\|Px\|$$ where $$P$$ is the orthogonal projection onto $$C^\perp$$. Since $$v\in C^\perp$$, your question is then whether there exists $$x\in F$$ which minimizes $$\|P(v-x)\|=d(v,Px)$$.
Now note that $$Pf_n=e_{2n}$$ so if we let $$x_N=\sum_{n=0}^N 2^{-n}f_n$$ then $$x_N\in F$$ for each $$N$$ and $$Px_N\to v$$. So, if there exists $$x\in F$$ which minimizes $$d(v,Px)$$, then we must have $$Px=v$$. Since the $$f_n$$ are an orthogonal basis for $$F$$, we can write such an $$x$$ as $$\sum c_n f_n$$ for scalars $$c_n$$, and then $$\langle x,e_{2n}\rangle=c_n$$. So if $$Px=v$$, we must have $$c_n=2^{-n}$$ for each $$n$$. But the sum $$\sum 2^{-n}f_n$$ does not converge since $$\|f_n\|>2^n$$, so no such $$x$$ exists.
I don't know what you mean by "the diagonal of $$\mathcal{H}$$", but by applying a translation and a unitary operator to everything in this example, you can get a similar example where $$D$$ is an arbitrary closed affine subspace of $$\mathcal{H}$$ whose dimension and codimension are both infinite.