Prove a space is Hilbert

Question

I got stucked in this problem and get no clue to solve this. Can any one please help me? Thanks

Suppose $X$ is an inner product space. If for every bounded linear function $f$, there exists $z \in X$ such that $$f(x) = <x, z>$$ for all $x \in X$. Prove that $X$ is a Hilbert space.

I can't find any way to relate the complete of space $X$ with the existence of $z$. Please help me solve this. I really appreciate.

Answer 1

Let $(x_k)_{k\geqslant 1}$ be a Cauchy sequence in $X$. Using Cauchy-Schwarz inequality, for each $u$, the sequence $(\langle x_k,u\rangle)_{k\geqslant 1}$ is a Cauchy sequence of real numbers. We define $$L(u):=\lim_{k\to \infty}\langle x_k,u\rangle.$$ Then $L$ is a continuous linear form, hence there is $y$ such that for all $u\in X$, $\langle x_k,u\rangle\to \langle y,u\rangle$.

So we have a candidate for the limit.

Fix $\varepsilon\gt 0$. We replace $x_k$ by $x_k-y$. There is $N=N(\varepsilon)$ such that $\lVert x_{k+j}-x_k\rVert\lt\varepsilon$ if $k\geqslant N$ and $j\geqslant 0$. We thus have $$|\langle x_k,u\rangle|\leqslant \varepsilon\lVert u\rVert+|\langle x_{k+j},u\rangle|,$$ hence $|\langle x_k,u\rangle|\leqslant \varepsilon\lVert u\rVert$ for each $u$ and $k\geqslant N(\varepsilon)$. Thanks to Cauchyness of the sequence $(x_k)_k$ we proved that the $N$ is independent of $u$. We thus have $$\lim_{k\to \infty}\sup_{\lVert u\rVert =1}|\langle x_k,u\rangle|=0,$$ hence $\lVert x_k\rVert\to 0$.

Answer 2

You can prove that $X'$ is a Hilbert Space. Then we can use the Riesz Representation Theorem on $X'$ and prove that $X$ is reflexive. Then $X$ is the same with $X''$ in the sense of isometry. Thus, we get $X$ is a Hilbert Space.

Answer 3

I will leave technical details to you and give you just a rough sketch of proof.

Take Cauchy sequence $z_1,z_2,\dots$, define functionals $f_n(x) = <z_n,x>$. Show that $f_n$ has a limit $f$(space of linear maps from space to complete space is complete). Then $z$ such that $f(x) = <z,x>$ is a limit of $z_1,z_2,\dots$.