I have the following question from Teschl's "Mathematical Methods in Quantum Mechanics":
A sesquilinear form is called bounded if $$\|s\|=\sup_{\|f\|=\|g\|=1}|s(f,g)|$$ is finite. Similarly, the associated quadratic form $q$ is bounded if $$\|q\|=\sup_{\|f\|=1}|q(f)|$$ is finite. Show $$\|q\|\le\|s\|\le2\|q\|.$$
There is a hint that says to use the parallelogram law and the polarization identity. Applying the polarization identity to $s$, we have $$s(f,g)=\frac14(q(f+g)-q(f-g)+iq(f-ig)-iq(f+ig)).$$ Now, the parallelogram law gives us $$q(f+g)=2q(f)+2q(g)-q(f-g)$$ and $$q(f+ig)=2q(f)+2q(ig)-q(f-ig)=2q(f)+2q(g)-q(f-ig).$$ So $$|s(f,g)|=|\frac14(2q(f)+2q(g)-2q(f-g)+2iq(f-ig)-2iq(f)-2iq(g))|$$ $$=|\frac12(q(f)+q(g)-q(f-g)+iq(f-ig)-iq(f)-iq(g))|.$$ If the $q(f-g)$ and $iq(f-ig)$ terms were absent, we could use the triangle inequality and then take the supremum to obtain the second inequality. However, $$-q(f-g)=-q(f)-q(g)+s(f,g)+s(g,f)$$ and $$iq(f-ig)=iq(f)+iq(g)+s(f,g)-s(g,f)$$ do not seem to cancel. Any help with this problem would be appreciated.
The first inequality follows straightforwardly from the fact that $s(f, f) = q(f)$. To prove the second inequality, we can note that if $\|q\| = A$, then $$q(f) \leq n^2 A$$ for all $\|f\| = n$. This follows from property that $q(f)$ scales quadratically when the argument is scaled, i.e. $q(af) = \|a\|^2 f$ for all $a \in \mathbb{C}$.
Using the first equation of your work $$s(f,g)=\frac14(q(f+g)-q(f-g)+iq(f-ig)-iq(f+ig))$$ we can apply the triangle inequality, which tells us that $$\|s(f, g)\| \leq \frac14 \left[ \|q(f + g)\| + \|q(f - g)\| + \|q(f - ig)\| + \|q(f + ig)\| \right]$$ Let us assume that $\|f\| = \|g\| = 1$. Then it follows that all of $f + g$, $f - g$, $f - ig$, and $f + ig$ have norm at most $\sqrt{2}$ (why?). From here, we can use what we discussed previously to conclude that each of the four terms in the parentheses above are at most $2 \|q\|$, from which the second inequality follows. $\square$