0
$\begingroup$

1) Consider the operator $A:L_2[0,1]\longrightarrow L_2[0,1]$ defined by $Ax(t)=\int_0^1t x(s)ds$. What is the adjoint of A?

2) Consider the operator $A:L_2[0,1]\longrightarrow L_2[0,1]$ defined by $Ax(t)=\int_0^1t x(t)dt$. What is the adjoint of A?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Why did you ask the same question twice? Also, what exactly have you tried so far? $\endgroup$ – Omnomnomnom Jun 15 '17 at 6:13
  • $\begingroup$ So one is an integral ds, the other is an integral dt. Still: the question doesn't quite make sense as written. Also, what have you tried? $\endgroup$ – Omnomnomnom Jun 15 '17 at 6:17
  • $\begingroup$ $f(x)$ appears nowhere in the right hand side of either of your expressions. $\endgroup$ – helloworld112358 Jun 15 '17 at 6:19
0
$\begingroup$

Hint: Currently, 2) is the only question which kind of makes sense as phrased, but my best interpretation is that $$ [Af](x) = \int_0^1 t f(t)\,dt $$ With that, $$ \langle Af,g \rangle = \int_0^1 \left(\int_0^1 t f(t)\,dt\right) \overline{g(s)}\,ds \\ = \int_0^1 \int_0^1 t \,f(t)\, \overline{g(s)}\,dt\,ds\\ = \int_0^1 \int_0^1 t \,f(t)\, \overline{g(s)}\,ds\,dt \\ = \int_0^1 f(t)\left(\overline{t\int_0^1 g(s)\,ds}\right)\,dt = \langle f, ??? \rangle $$

$\endgroup$

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.