Need a help in understanding example(1) on Riesz representation theorem.

Question

The author said:

"A functional $F$ on $L^{2}([a,b])$ is bounded and linear iff there exists a $g \in L^{2}([a,b])$ such that $$F(f) = \int_{a}^{b} f(t) \bar{g}(t) dt,$$

for all $f \in L^{2}([a,b])$. In this case $||F|| = ||g||.$"

**$Q_{1}$**I do not understand why the functional $F$ is bounded and linear iff the above condition is satisfied ?

**$Q_{2}$**Also I do not understand why in this case $||F|| = ||g||$?

Could anyone explain this for me please?

Thanks!

Answer 1

If $F$ is linear and bounded then such a $g$ existes because that's what the Riesz representation theorem says.

On the other hand, if you take $g\in L^2\bigl([a,b]\bigr)$ and then define $F$ as you did, it is easy to see that $F$ is linear. If $\|f\|=1$, then\begin{align}\bigl|F(f)\bigr|&=\left\|\int_a^bf(x)\overline{g(x)}\,\mathrm dx\right\|\\&\leqslant\|f\|.\|g\|,\end{align}by Cauchy-Schwarz. Therefore, $F$ is bounded and $\|F\|\leqslant\|g\|$. In order to see that $\|F\|=\|g\|$, there are two possibilities:

• $g\equiv0$: then $\|F\|=\|g\|=0$;
• $g\not\equiv0$: then $\left\|\frac g{\|g\|}\right\|=1$ and $\left|F\left(\frac g{\|g\|}\right)\right|=\|g\|$. Therefore, $\|F\|\geqslant\|g\|$. But we already saw that $\|F\|\leqslant\|g\|$.