Q:

Bounding the Riesz Isomorphism

I'm working through the details of verifying various facts about the Riesz Isomorphism between a normed vector space $(E, || \cdot ||_E)$ (with an inner product) and its dual $E'$. One of the things that I want to show is that the map 
$f_y(x) = (x | y)$ is bounded and, in fact, $||f_y||_{E'} \leq ||y||_E$ where $||\cdot||_{E'}$ denotes the operator norm on $E'$. I believe that I have this worked out but would appreciate a second opinion as I always feel inept when attempting proofs that involve estimation!
First, I claim $f_y$ is bounded (by $||y||_E$) because 
$$
|f_y(x)| = |(x|y)| \leq  ||y||_E \cdot ||x||_E$$
where the last inequality follows from Cauchy-Schwarz. Therefore, the operator norm can be meaningfully applied to $f_y$. Furthermore,
$||f_y||_{E'}$
$= \sup\{|f_y(x)| : ||x||_E \leq 1 \}$
$= \sup\{|(x|y)| : ||x||_E \leq 1 \}$
$
\leq \sup\{ ||x||_E \cdot ||y||_E \ : ||x||_E \leq 1 \}
$
$
=||y|| \sup\{||x||_E : ||x||_E \leq 1 \}
$
$
= ||y||_E
$

Here, the fourth line follows again from Cauchy-Schwarz.
So, my question is, does this proof work or are there gaps in my reasoning?

A:

It is correct, but it seems that you did the same thing twice. To save some time, it might be useful to know the following. I think your definition of the operator norm is
$\|L\|_{op}:=\sup\{\|Lv\|:v\in E\text{ and }\|v\|\leq 1\}$ .
Try to prove that this is the same as
$\inf\{C\geq 0:\|Lv\|\leq C\|v\| \text{ for all }v\in E\}$ .
This can save some time. For example, you first showed that $f_y$ is bounded by estimating
$|f_y(x)|\leq \|y\|\|x\|$.
Now you see immediately that $\|f_y\|_{op}$ is at most $\|y\|$, as $\|y\|$ occurs in the set over which you take the infimum.

