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Let $X$ be an inner product space. For any bounded linear functional $f$ on X, there exists a unique $x_f \in X$ s.t. for any $x \in X$, $f(x)=\langle x, x_f\rangle$, and $\|f\|=\|x_f\|$. Show that $X$ is a Hilbert space.

Remember: It is true that Riesz's representation theorem doesn't hold for an incomplete inner product space, but we cannot use it to solve this problem because they are completely different.

Here's my idea:

Let $\{x_n\}$ be a Cauchy sequence. For a given functional $f$, by the fact that $|\langle x_n, x_f\rangle| ≦ \|x_f\|\|x_n\|$, we have $\{\langle x_n, x_f\rangle\}$ is a Cauchy sequence. However, we can't immediately find a $x$ s.t. $\{\langle x_n, x_f\rangle\} \rightarrow \{\langle x, x_f\rangle\}$.

Any hint will be most welcomed.

Calvin Khor
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user444628
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  • Hint: If $X$ is an inner product space (complete or not), then $X^{*}$ is a complete normed space. –  Jan 02 '18 at 06:51

3 Answers3

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For your Cauchy sequence $x_n$, consider them as linear functionals $f_{x_n}(x) := \langle x_n,x\rangle $. The dual of the normed space $X$ is complete, so $f_{x_n}$ have a limit, $f\in X'$. By hypothesis, there is some $x_f\in X$ such that $f(x) = \langle x_f,x \rangle$. Since $f_{x_n} - f$ is represented by the element $x_n - x_f \in X$, we also know by the given hypothesis that $$ \|x_n - x_f \|_X = \| f_{x_n} - f\|_{X'} \to 0.$$

Calvin Khor
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Define the linear functional $\newcommand{\ang}[1]{\left\langle{#1}\right\rangle}f : y\mapsto \lim_n \ang{y,x_n}$ (using the fact that $\ang{y,x_n}$ is a Cauchy sequence as you noted). Then let $x_f$ be such that $f(y) = \ang{y,x_f}$ by the assumption. Then we just need to show that $x_n \to x_f$.

Now consider $x_n-x_f$, $$\ang{x_n-x_f,y}=\lim_{m\to\infty} \ang{x_n-x_m,y}\le \lim_{m\to\infty} \|x_n-x_m\|\|y\|,$$ so since $x_n$ is Cauchy, $\ang{x_n-x_f,y}\to 0$ for any $y$, in particular, for $x_n$, and $x_f$. Thus for large enough $n$, $$\|x_n-x_f\|^2 = \ang{x_n-x_f,x_n-x_f} \le |\ang{x_n-x_f,x_n}| + | \ang{x_n-x_f,x_f}|\le \epsilon$$ for any $\epsilon>0$. Hence $x_n\to x_f$ as desired.

jgon
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Using the assumption you can establish an antilinear isometric isomorphism between $X^*$ and $X$.

Define $A : X \to X^*$ as $$Ax = f_x = \langle \cdot, x\rangle = v \mapsto \langle v, x\rangle$$

for all $x \in X$. The assumption gives us that $A$ is bijective.

$A$ is antilinear:

$$A(\alpha x + \beta y) = \langle \cdot, \alpha x + \beta y\rangle = \overline{\alpha} \langle \cdot, x\rangle + \overline{\beta}\langle \cdot, y\rangle = \overline{\alpha} Ax + \overline{\beta}Ay$$

Finally, $A$ is isometric:

$$\|Ax\| = \|f_x\| = \|x\|$$

Now just use that (linear or antilinear) isometric isomorphisms preserve completeness:

$$\|f_{x_m} - f_{x_n}\| = \|Ax_m - Ax_n\| = \|A(x_m - x_n)\| = \|x_m - x_m\|$$

So one sequence is Cauchy/convergent if and only if the other one also is.

Therefore, since $X^*$ is always complete, we conclude that $X$ is also complete.

mechanodroid
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