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

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



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 $y$?

  • $\begingroup$ 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

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