Let $\mathcal{H}$ be a Hilbert space, and $\mathcal{H}^*$ its topological dual space (the space of continuous linear forms on $\mathcal{H}$). The exists a conjugate-linear isometry between these two Hilbert spaces, so it stands to reason that

$$\langle f_\vec{v}|f_\vec{w}\rangle \space = \langle\vec{w}|\vec{v}\rangle$$

would be the inner product on $\mathcal{H}^*$, and thus

$$\lVert f_\vec{v}\rVert = \sqrt{\langle\vec{v}|\vec{v}\rangle} = \lVert\vec{v}\rVert.$$

But I've been taught that the norm of a linear form on a Hilbert space is

$$\lVert f_\vec{v}\rVert = \mathrm{sup}\left\{\frac{|f_\vec{v}(\vec{x})|}{\lVert\vec{x}\rVert}\space|\space \vec{x}\in\mathcal{H}\smallsetminus\{0\}\right\}.$$

Are these two norms equivalent? How could I prove it?

  • 1
    $\begingroup$ By Cauchy's inequality $|f_v(x)|=|\langle x,v\rangle|\leq \|v\|\|x\|$. Therefore $\|f_v\|\leq \|v\|$. But also $\|f_v\|\geq \frac{|f_v(v)|}{\|v\|}=\frac{|\langle v,v\rangle|}{\|v\|}=\|v\|$. $\endgroup$
    – logarithm
    May 13 '19 at 12:07
  • $\begingroup$ You mean norm squared $\endgroup$
    – user515599
    May 13 '19 at 12:09
  • $\begingroup$ @badatmath true, thank you. $\endgroup$
    – TeicDaun
    May 13 '19 at 12:12

The user @logarithm already answered correctly in their comment. I'm writing here an answer for the question to be closed.

The key is to use Cauchy(-Schwarz)'s inequality, $$|\langle\vec{v}|\vec{x}\rangle| \leq \lVert\vec{v}\rVert\lVert\vec{x}\rVert.$$ One way to do it is starting off with the second definition of $\lVert f_{\vec{v}}\rVert$ and using the fact that Cauchy's inequality is saturared if and only if $\vec{x}=\lambda\vec{v}$. $$\lVert f_{\vec{v}}\rVert = \mathrm{sup}\left\{\frac{|f_\vec{v}(\vec{x})|}{\lVert\vec{x}\rVert}\right\} = \mathrm{sup}\left\{\frac{|\langle\vec{v}|\vec{x}\rangle|}{\lVert\vec{x}\rVert}\right\} = \frac{|\langle\vec{v}|\lambda\vec{v}\rangle|}{\lVert\lambda\vec{v}\rVert} = \frac{|\lambda|}{|\lambda|}\frac{|\langle\vec{v}|\vec{v}\rangle|}{\lVert\vec{v}\rVert} = \frac{\lVert\vec{v}\rVert^2}{\lVert\vec{v}\rVert} = \lVert\vec{v}\rVert$$ as we wanted to prove.

It's maybe easier and more elegant to do what @logarithm did, though. We can use Cauchy's inequality directly to see that $|f_\vec{v}(\vec{x})|\leq \lVert\vec{x}\rVert\lVert\vec{v}\rVert \implies \lVert f_\vec{v}\rVert\leq\lVert\vec{v}\rVert$, but the definition of $\mathrm{sup}\{\}$ implies, trying $\vec{x}=\vec{v}$, that $\lVert f_\vec{v}\rVert\geq\lVert\vec{v}\rVert$. Thus, we have $\lVert f_\vec{v}\rVert = \lVert\vec{v}\rVert$ as we wanted to prove.


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