$f_k := e_k + e_{k+1}$ complete, but $e_1 \neq \sum_{k=1}^{\infty} \alpha_k f_k$ for any $\alpha_k$

Question

Let $\mathcal H$ be a Hilbert space and $\{e_k\}_{k=1}^{\infty}$ a orthonormal basis for $\mathcal H$. Consider $f_k := e_k + e_{k+1}$, $k \in \mathbb N$.

Claim: $\{f_k\}_{k=1}^{\infty}$ is complete, i.e. $M:= \overline{\operatorname{span}(\{f_k\}_{k=1}^{\infty})} = \mathcal H$.

Proof: Assume not. Then there exists $f \in M^⊥ \setminus \{0\}$ such that $\langle f, f_k \rangle = 0$ $\forall k \in \mathbb N$. Then, $\langle f, e_k \rangle = - \langle f, e_{k+1} \rangle$ $\forall k$ which implies that $|\langle f, e_k \rangle|$ is a constant. Since $\sum_{k=1}^{\infty} |\langle f, e_k \rangle|^2 = ||f||^2 \lt \infty$, we have that $f = 0$. But this is a contradiction to the assumption and thus $\{f_k\}_{k=1}^{\infty}$ is complete.

On the other hand, one can see that for example $e_1$ can not be written as $e_1 = \sum_{k=1}^{\infty} \alpha_k f_k$ for any choice of the coefficients $\{\alpha_k\}_{k=1}^{\infty} \subset \mathbb K$.

If I haven't made a mistake, this seems like a contradiction. Opinions?

Answer 1

Note $$\| e_1-\sum_{k=1}^n \alpha_k f_k \|^2=\| (1-\alpha_1)e_1+\sum_{k=2}^{n}(\alpha_{k-1}+\alpha_k)e_k+\alpha_ne_{n+1} \|^2\\=(1-\alpha_1)^2+\sum_{k=2}^{n}(\alpha_{k-1}+\alpha_k)^2+\alpha_n^2$$

Now, setting $\alpha_k=(-1)^k \frac{n-k}n$ we get

$$\| e_1-\sum_{k=1}^n \alpha_k f_k \|^2=\| (1-\alpha_1)e_1+\sum_{k=2}^{n}(\alpha_{k-1}+\alpha_k)e_k+\alpha_ne_{n+1} \|^2\\=\frac{1}{n^2}+\frac{n-1}{n^2}+ \frac{1}{n^2}=\frac{n+1}{n^2} \to 0$$

Answer 2

Hmm. "Seems like" is a little fuzzy - it may well seem like a contradiction, but it's not.

It's not a contradiction: Well, I can't prove it's not a contradiction, all I can say is you haven't actually demonstrated anything contradictory. It would be a contradiction if the $f_k$ were orthogonal, but they're not; this is a very special property of orthogonal series.

"Seems like": It may well "seem like" if the closed span is dense then everything should be equal to some series $\sum\alpha_kf_k$. But that's simply not so.

A familiar example of the same phenomenon, that you may have never noticed: Say $X=C([0,1])$ with the sup norm. Weierstrass says that the closed span of the monomials $t^k$ is dense. But it certainly does not follow that every continuous function has the form $\sum\alpha_kt^k$; only analytic functions can be written that way.