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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 ?

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1 Answer 1

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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.

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