Given a countable orthonormal set in a Hilbert Space, a convergent sequence in the span of the set implies convergence of coefficients.

Question

The problem is from Naylor's Linear Operator Theory in Engineering and Science

(Section 5.17, Exercise 19) Let $B = \{x_1, x_2, . . . \}$ be an orthonormal set in a Hilbert space $H$. Then let ${z_n}$ be a convergent sequence in $H$, where each $z_n$ is a finite linear combination of points in $B$, that is, $z_n = α_{n1}x_1 + α_{n2}x_2 + · · · + α_{nk_n} x_{k_n}$ . Show that for each $j = 1, 2, . . .$ the sequence $\{α_{nj}\}$ is convergent, that is, $$α_j = \lim_{n→∞}α_{nj}$$ and that $$\lim_{n \to \infty}z_n = \lim_{n \to \infty} \sum_{j=1}^{k_n}α_{nj}x_j = \sum_{j=1}^{∞} α_jx_j .$$

I'm been stuck on this for a while since I can't say $B$ is a basis for $H$. Also I know that since $z_n$ is convergent, $\sum_{j=1}|α_{nj}|^2<\infty $, but I don't see where I can use this.

I was thinking of using the convergence of $z_n$ to some $z \in H$ to get the convergence of $α_{nj}$.

Answer 1

Note, for all $j$, that $\langle\cdot,x_j\rangle:H\to\mathbb{K}$ is continuous. Furthermore, for all $n$ and $j$, we have $\langle z_n,x_j\rangle=\alpha_{nj}$, because $B$ is orthonormal. Hence, since $z_n$ converges, so does $\alpha_{nj}$, for all $j$, to some $\alpha_j$.

Furthermore, if $z$ is the limit of $z_n$, from the continuity we also get $\langle z,x_j\rangle=\alpha_j$, for all $j$. For all $n$ and $j$ we get $\langle z-z_n,x_j\rangle=\alpha_j-\alpha_{jn}$. Since $B$ is orthonormal, this gives $\sum_{j=1}^\infty(\alpha_j-\alpha_{nj})^2\leq\|z-z_n\|^2$, for all $n$.

In particular, the sum converges for all $n$, and since the sequence $(\alpha_{nj})_{j\geq1}$ is finite, this implies that $\sum_{j=1}^\infty\alpha_j^2$ also converges. Since $B$ is orthonormal, $\sum_{j=1}^\infty\alpha_jx_j$ also converges to some vector $v$.

Finally, for all $n$ we have $\|v-z_n\|=\|\sum_{j=1}^\infty(\alpha_j-\alpha_{jn})x_j\|=\sum_{j=1}^\infty(\alpha_j-\alpha_{nj})^2$, which tends to zero as $n$ tends to infinity. Since $z_n$ converges to both $z$ and $v$, we have $z=v=\sum_{j=1}^\infty\alpha_jx_j$.

Answer 2

Let $z_n \to z$. Note that $\alpha_{nj}=\langle z_n , x_j \rangle$. Hence $ \alpha_{nj} \to \langle z, x_j \rangle$ for each $j$ as $ n \to \infty$. Thus $\alpha_j=\langle z, x_j \rangle$. Now $z$ belongs to the closed subspace spanned by $x_1,x_2,...$. The sequence $(x_j)$ is an orthnormal basis for this space. Hence we can write $z=\sum_j a_j x_j$. Now $\langle z , x_j \rangle =a_j$ so $\alpha_j=a_j$ for all $j$. This finishes the proof.