Let $X$ be an inner product space. Show that $X$ is a Hilbert space if and only if for each continuous linear functional $L$ on $ X$, there exists $z \in X$ such that $$ L(x) = \langle x,z \rangle $$
Here, one part is exactly the Riesz representation theorem. How can I prove the converse result? That is, If for each continuous linear functional $ L$ on $ X$,there exists $ z\in X$ such that $ L(x)=\langle x,z\rangle $ then $ X $ is a Hilbert space.Any Help is appreciated.
Hint: Suppose $x_n$ is a Cauchy sequence in $X$.
Consider the linear functional $L(x) = \lim_n \langle x, x_n \rangle$.
by the hints we know that $\langle x, x_n \rangle$ is a Cauchy sequence in $\mathbb{K}$. Hence it converges. We know that $L(x) = \langle x, z \rangle \rightarrow \langle x,z \rangle = \lim_n \langle x, x_n \rangle$. By choosing $L(x_n-z)$ we get $x_n = z$. So every Cauchy sequence in $X$ converges so $X$ is a Hilbert space.
Note that the limit $\langle x,x_n \rangle $, converge because
$$ | \langle x,x_n \rangle- \langle x,x_m \rangle |= |\langle x,x_n-x_m \rangle|\leq\|x\|\|x_n-x_m\| $$
Then $x_n$ cauchy implies that $\langle x,x_n \rangle $ cauchy( in $\mathbb R$ or $\mathbb C$)
How above define a continuous linear functional $L(x)=\lim\limits_{n } \langle x,x_n \rangle $.
By hypothesis there is a $z\in X$ such that $L(x)=\langle x,z \rangle $. Even we know in advance that X is Hilbert then we will conclude only that $x_n$ converge to $z$ weakly.
This is not a answer of course. Is a comment that does not fit in the place of comment.
@RobertIsrael,@ccc Can you give another hint?
This is an old question but since none of the answeres are complete I am giving a complete solution. Let $x_{n}$ be a Cauchy sequence
First notice that $\langle y,x_{n}\rangle$ will always be a Cauchy sequence in $\Bbb{K}$ for each $y\in H$ as $|\langle y,x_{n}\rangle-\langle y,x_{m}\rangle|\leq ||y||\cdot||x_{n}-x_{m}||$ .
Thus $\lim_{n\to\infty} \langle y,x_{n}\rangle$ always exists.
As per Robert's answer define for each $y\in H$ , $L(y)=\lim_{n\to\infty}\langle y,x_{n}\rangle$ . Then $L$ is a linear functional and it is also bounded due to Cauchy-Schwartz inequality as $||x_{n}||$ is a bounded sequence.
Thus $L(\cdot)=\langle \cdot,z\rangle$ for some $z\in H$ by hypothesis.
Now let $\epsilon>0$ and $N$ be such that $||x_{n}-x_{m}||<\epsilon$ for all $n,m\geq N$
Then for a fixed $n\geq N$ consider the continuous functional $T(y)=\langle y,\frac{x_{n}-z}{||x_{n}-z||}\rangle$ . Then $||T||=1$ and $||T(x_{n}-z)||=||x_{n}-z||$ . Then notice that as $\langle y,x_{m}\rangle \to \langle y,z\rangle$ for each $y$ , we have $\displaystyle\lim_{m\to\infty}\langle \frac{x_{n}-z}{||x_{n}-z||}, x_{m}\rangle= \langle \frac{x_{n}-z}{||x_{n}-z||}, z\rangle $ which also means that $\displaystyle\lim_{m\to\infty}\langle x_{m},\frac{x_{n}-z}{||x_{n}-z||} \rangle= \langle z,\frac{x_{n}-z}{||x_{n}-z||}\rangle $
(Notice that we are taking limit wrt $m$ and not $n$ )
Then for $m\geq n\geq N$ we have
$\begin{align}||x_{n}-z||=||T(x_{n}-z)||&=||T(x_{n}-x_{m})+T(x_{m}-z)||\\ &\leq ||T||\cdot||x_{n}-x_{m}||+\bigg\vert\langle x_{m},\frac{x_{n}-z}{||x_{n}-z||}\rangle-\langle z,\frac{x_{n}-z}{||x_{n}-z||}\rangle\bigg\vert\\ &\leq \epsilon + \bigg\vert\langle x_{m},\frac{x_{n}-z}{||x_{n}-z||}\rangle-\langle z,\frac{x_{n}-z}{||x_{n}-z||}\rangle\bigg\vert \end{align}$
Now we can let $m\to\infty$ so that the term above is also $<\epsilon$ .
So $||x_{n}-z||\leq 2\epsilon$ for all $n\geq N$ .
Note that what Robert's answer does is show that $x_{n}\xrightarrow{w} z$ . What we have done is show that Cauchy+weak convergence in a normed space is equivalent to Convergence. Now essentially we are producing functionals $f_{n}$ of operator norm $1$ such that $|f_{n}(x_{n}-z)|=||x_{n}-z||$. In a more general space(not necessarily complete) , you'll need Hahn Banach Theorem to do so. Now we are using the Cauchyness of $x_{n}$ and the fact that $f_{n}(x_{m})\xrightarrow{m\to\infty} f_{n}(z)$ to conclude that $x_{n}\to z$ .
