I have a question from Hilbert space theory.
Let $\mathcal H$ be a separable Hilbert space. Suppose $\{x_n\}$ is a sequence of vectors in $\mathcal H$ such that $$\inf\{\|x_m - x\|: x\in \overline{\operatorname{span}\{x_k\}_{k\ne m}}\}\ge \beta > 0$$ for every $m\ge 1$, for a fixed $\beta$. Show that there exists a bounded sequence $\{y_n\}$ in $\mathcal H$ such that $\langle x_m,y_n\rangle = \delta_{m,n}$ for all $m,n\ge 1$.
Some work: I have shown that one can find $\{y_n\}$ satisfying the orthogonality relation $\langle x_m,y_n\rangle = \delta_{m,n}$. To do this, define $S_m := \overline{\operatorname{span}\{x_k\}_{k\ne m}} \subset \mathcal H$. Of course, $\mathcal H = S_m \oplus S_m^\perp$ and $S_m^\perp \ne \{0\}$. We can choose any $z_1 \in S_1^\perp\setminus\{0\}$ such that $\langle x_1, z_1\rangle \ne 0$. Normalizing $z_1$ with $\overline{\langle x_1, z_1\rangle}$, we get $y_1\ne 0$ with $\langle x_1,y_1\rangle = 1$ and $\langle x_1,y_n\rangle = 0$ for all $n\ge 1$. We can repeat this process countably many times to get a sequence $\{y_n\}$ such that $\langle x_m, y_n\rangle = \delta_{m,n}$. Now, I'm not sure how to produce $C > 0$ such that $\|y_n\| \le C$ for all $n\ge 1$. Thanks!
You are on the right track - though you could just directly write down $y_n = \alpha_nP_{S_n^\perp}(x_n)$, where $\alpha_n$ is the normalization constant given by $\alpha_n = \frac{1}{||P_{S_n^\perp}(x_n)||^2}$. Since $||P_{S_n^\perp}(x_n)|| = \inf\{||x_n - x||: x \in S_n\} \geq \beta$, we have $||y_n|| = \alpha_n||P_{S_n^\perp}(x_n)|| = \frac{1}{||P_{S_n^\perp}(x_n)||} \leq \frac{1}{\beta}$ for all $n$.