# Inner product on dual of separable inner product space

Let $$H$$ be a separable inner product space and let $$H'$$ denote the dual of $$H$$ (the set of bounded linear functionals).
Consider $$(u_n)$$ a countable family of orthonormal vectors in $$H$$ with dense linear span.

For $$f,g \in H'$$ define $$\langle f,g \rangle = \sum_{j=1}^\infty \overline{f(u_j)}g(u_j)$$.
Then:

• $$\langle \cdot,\cdot \rangle$$ is an inner product on $$H'$$
• $$H'$$ is is a separable Hilbert space w.r.t. to $$\langle \cdot,\cdot \rangle$$
• the norm induced by $$\langle \cdot,\cdot \rangle$$ is the usual operator norm.

I have trouble even proving that $$\langle f,g \rangle$$ is well-defined... Here what I've tried. For any $$N\geq 1$$, $$\sum_{j=1}^N |\overline{f(e_j)}g(e_j)| \leq \left(\sum_{j=1}^N|f(e_j)|^2 \right)^{1/2} \left(\sum_{j=1}^N|g(e_j)|^2 \right)^{1/2}$$

and $$\sum_{j=1}^N|f(e_j)|^2 = \|\sum_{j=1}^N f(e_j) e_j \|^2,$$

but why should $$\sum_{j\geq 1} f(e_j) e_j$$ converge ?

Another line of thought: $$\sum_{j=1}^N|f(e_j)|^2 \leq \|f\| \sum_{j=1}^N \|e_j\|^2 = \|f\| \Big |\Big |\sum_{j=1}^N e_j \Big |\Big |^2$$

but why should $$\sum_{j\geq 1}e_j$$ converge ?

For any $$x = \sum x_i e_i$$ in $$H$$, we have $$|g(x)|\le \|g\| \|x\|$$, which means
$$\tag{1} \left| \sum_{i=1}^\infty x_i g(e_i)\right| \le \| g\| \|x\|.$$
Now for each $$N$$, let $$x_N = \overline{g(e_1)}x_1 + \cdots \overline{g(e_n)} e_n$$. Then from (1),
\begin{align} \sum_{i=1}^N |g(e_i)|^2 &= \left| \sum_{i=1}^N \overline{g(e_i)} g(e_i) \right| \\ &\le\| g\| \sqrt{\sum_{i=1}^N |g(e_i)|^2}\\ \Rightarrow\sqrt{\sum_{i=1}^N |g(e_i)|^2} &\le \|g\|. \end{align}
Thus together with your calculations, $$\langle f, g\rangle \le \|f\| \cdot \|g\|$$ and so the inner product is well defined.