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Let $E = L^2 (0, 1)$. Given $u ∈ E$, set $Tu(x)=\int_0^x u(t)dt$.

Find $T^*$.

Solution says only $(u, T^* v) = \int_t^1 v(x)dx$.

I dont understand. adjoint operator is defined by $(Tu,v)=(u,T^*v)$. How do we get the formulae above?

Edit

Wew have $(Tu,v)=(u,T^*v)=\int_0^1 (\int_0^x u(t)dt)v(x)dx=\int_0^x u(t)(\int_0^1 v(x)dx)dt$ by interchanging the integrals but It looks like there is a variable change to do but I can't see where.

Thank you for any help/hints.

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  • $\begingroup$ The sentence beginning with "Solution says" does not make sense. How is the action of $T^*v$ on $u$ independent of $u$? $\endgroup$ – uniquesolution May 5 at 8:59
  • $\begingroup$ @uniquesolution I am not sure why it should depend on $u$. There are constant operator that does not depend on any variable, right? I am trying to write $(u,T^*v)$ as an integral as you can see in my edit, but I don't understand what it the interval on which we should integrate $\endgroup$ – PerelMan May 5 at 9:18
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    $\begingroup$ When you changed the order of integration you didn't adjust the limits appropriately (your outer integrand limits shouldn't depend on the inner variable). The expression $\int_0^1 \int_0^x dt dx$ can be interpreted as integrating over the the triangle bounded by $x=0$, $t=1$, and $t=x$. When reordering the limits should be $\int_0^1 \int_x^1 dx dt$. The limits describe the same triangle, but with the variables flipped. $\endgroup$ – Zorngo May 6 at 4:14
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$$ \langle Tu,v\rangle = \int_0^1 \int_0^x u(y)dy\; \overline{v(x)}dx \\ = \int_0^1\int_0^1\chi_{[0,\infty)}(x-y)u(y)dy\,\overline{v(x)}dx \\ = \int_0^1\int_0^1\chi_{[0,\infty)}(x-y)\overline{v(x)}dx\, u(y)dy \\ = \int_0^1 u(y) \overline{\int_y^1 v(x)dx} dy = \langle u,T^*v\rangle $$ Therefore $$ T^*v = \int_y^1 v(x)dx. $$

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