0
$\begingroup$

I have a functional $f: L^2[-1,1]\to\mathbb{C}$, how do I find a $v$ such that $f(x)=\langle x,v\rangle$ for all $x$? Since $L^2$ is a Hilbert space, I know the Riesz Representation theorem holds but I'm not sure what form $v$ is supposed to take... The functional is

$$ f(x) = i\int_{-1}^{1} x(t) dt + \int_{0}^{1} tx(t)dt $$

Thanks

$\endgroup$
2
$\begingroup$

What you want to end up with is something of the form

$$f(x) = \int_{-1}^1 x(t)\overline{y(t)}\,dt.$$

This is the usual inner product on $L^2([-1,1])$. We can rewrite your expression slightly as

$$f(x) = \int_{-1}^1 i\,x(t)\,dt + \int_{-1}^1 \chi_{[0,1]}(t)t\,x(t)\,dt.$$

Do you see now how to get your $y$?

$\endgroup$
  • $\begingroup$ Ok great, thanks. I see how to get y but I'm not sure I understand how you rewrote the second term in the expression to change the limits? $\endgroup$ – user281547 Oct 18 '15 at 17:07
  • $\begingroup$ Do you know what the notation $\chi_{[a,b]}$ means? $\endgroup$ – Cameron Williams Oct 18 '15 at 17:28
  • $\begingroup$ No, I have not seen this before. $\endgroup$ – user281547 Oct 18 '15 at 19:09
  • $\begingroup$ Okay then $$\chi_{[a,b]}(x) = \begin{cases} 1 & x\in[a,b] \\ 0 & x\not\in[a,b]\end{cases}.$$ $\endgroup$ – Cameron Williams Oct 18 '15 at 19:15
  • $\begingroup$ You could write it in terms of the Heaviside step function if you're so inclined. $\endgroup$ – Cameron Williams Oct 18 '15 at 19:18

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.