# 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?

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Why would there be an inner product on $E$? – Davide Giraudo Jun 5 '11 at 21:39
I should have stated that E is assumed to have an inner product; fixed. – ItsNotObvious Jun 5 '11 at 21:47
Looks fine to me. – Mark Jun 5 '11 at 21:49
A totally minor note on language. You said:  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.  I think that the last inequality is (the abstract version of) Cauchy-Schwarz, it does not follow from Cauchy-Schwarz. Apart from that, everything is fine to me too. – Giuseppe Negro Jun 5 '11 at 22:55
@dissonance: I think in general $P$ implies $P$, so the CS inequality follows from the CS inequality, albeit for trivial reasons. – wildildildlife Jun 6 '11 at 14:28

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.

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Thanks for the tip – ItsNotObvious Jun 6 '11 at 15:02