2
$\begingroup$

I have been doing some research on fixed point theorems, and they have brought me around to papers from the 1960s in functional analysis in Banach spaces.

I think that today it is common practice to use either $\langle\cdot,\cdot\rangle$ or $(\cdot,\cdot)$ to mean an inner product in a Hilbert space. However, in these papers, they use the latter notation to mean something different: here, the first argument is a bounded conjugate linear functional from $V$, the second argument is in $V$, and $(f,x)=f(x)$.

Is there some reason that these two seemingly disparate concepts have the same notation? I do not know a lot of the language in linear algebra so I am a bit out of my league in terms of looking things up.

$\endgroup$
1
  • 2
    $\begingroup$ It is actually still common to use $(f,x)=f(x)$ even today. This notation better illustrates the bilinear nature of the map $V^*\times V\to \mathcal{k}$. It is more symmetric and illustrates the duality between a banach space and its dual. $\endgroup$ – Hui Yu Jul 7 '13 at 1:41
2
$\begingroup$

If $V$ is a Banach space and $V^*$ its is dual, the duality operator $(f,v)=f(v)$ is a bilinear operator. Moreover if $V$ is a Hilbert space you can identify each $f\in V^*$ with some $v_f$ in $V$. In that case it happens that $(f,v) =\langle v_f,v\rangle$.

$\endgroup$

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.