A property of inner products

Question

Let $H$ be an inner product space. Show that if $\sum_ku_k$ converges in $H$, then

$\langle\sum_ku_k,g\rangle=\sum_k\langle u_k,g\rangle$, where $ u_k,g\in H$.

This propery seems to follow directly from the axioms of inner products, but i am not sure if i am missing something. How would you prove this result?

Answer 1

Let $u=\displaystyle\sum_{k}u_{k}$ and $s_{n}=\displaystyle\sum_{k=1}^{n}u_{k}$, then $s_{n}\rightarrow u$ in $H$ and the continuity of first coordinate shows that \begin{align*} \left<u,g\right>=\lim_{n\rightarrow\infty}\left<s_{n},g\right>=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left<u_{k},g\right>=\sum_{k}\left<u_{k},g\right>. \end{align*}

For the convergence: $\left|\left<s_{n}-u,g\right>\right|\leq\|s_{n}-u\|\|g\|\rightarrow 0$ and we use Squeeze Theorem to conclude.

Answer 2

Set $\sum_{k = 1}^{\infty} u_k = v$. Let $\epsilon > 0$. Assume for convenience that $g \neq 0$, as otherwise the result is trivial.

By the convergence of the sequence $( \sum_{k = 1}^{N} u_k )_{N \in \mathbb{N}} \to v$, we know there exists $N_0$ such that $ \left\| v - \sum_{k = 1}^{N} u_k \right\| < ( \epsilon / \| g \| )^2$ for $N \geq N_0$. We know by the Cauchy-Schwarz inequality that if $N \geq N_0$, then \begin{align*} \left| \left< v - \sum_{k = 1}^{N} u_k , g \right> \right|^2 & \leq \left\| v - \sum_{k = 1}^{N} u_k \right\|^2 \cdot \| g \|^2 \\ & < (\epsilon / \| g \|)^2 \cdot \| g \|^2 \\ & = \epsilon^2 \\ \Rightarrow \left| \left< v - \sum_{k = 1}^{N} u_k , g \right> \right| & < \epsilon . \end{align*} This establishes the convergence we wanted.