1
$\begingroup$

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?

$\endgroup$
  • 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 $\endgroup$ – why May 13 '19 at 12:09
  • $\begingroup$ @badatmath true, thank you. $\endgroup$ – TeicDaun May 13 '19 at 12:12
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.