I would like to find the adjoint operator of $$ T\colon L^2([0,1])\to H^1([0,1]),\quad x\mapsto\int\limits_0^t x(s)\, ds. $$ Here $H^1([0,1])$ is the Sobolev space $W^{1,2}([0,1])$.

I tried to find this adjoint operator and after some time I came to the conclusion that this operator is given by $$ x\mapsto \int_t^1 x(s)\, ds + \overline{x'(t)}. $$ Is this correct? Because the calculation is rather long, I did not write it down here. But if it may be necessary to see the calculation, then i will add it of course.


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    $\begingroup$ It looks more or less okay (I think I see where you get everything), but I am not sure about the complex conjugation over the $x'(t)$. Perhaps it would be better if you showed the computation. $\endgroup$ – Willie Wong Jan 11 '13 at 12:20
  • $\begingroup$ okay, i added it. $\endgroup$ – math12 Jan 11 '13 at 12:46

So my result is wrong?

My calculation now is:

$\langle Tf,g\rangle_{H^1}=\langle Tf,g\rangle_{L^2} + \langle (Tf)',g'\rangle_{L^2}$

$=\int\limits_0^1 (Tf)(x)\overline{g(x)}\, dx+\int\limits_0^1 (Tf)'(x)\overline{g'(x)}\, dx$

$=\int\limits_0^1\int\limits_0^x f(s)\, ds\overline{g(x)}\, dx+\int\limits_0^1 f(x)\cdot \overline{g'(x)}\, dx$

$=\int\limits_0^1\int\limits_s^1 f(s)\overline{g(x)}\, dx\, ds+\int\limits_0^1 f(x)\overline{g'(x)}\, dx$

$=\int\limits_0^1 f(s)\int\limits_s^1\overline{g(x)}\, dx\, ds+\int\limits_0^1 f(x)\overline{g'(x)}\, dx$

$=\int\limits_0^1 \left(f(s)\int\limits_s^1\overline{g(x)}\, dx+f(s)\overline{g'(s)}\right)\, ds$

$=\int\limits_0^1 f(s)\cdot \left(\int\limits_s^1\overline{g(x)}\, dx+\overline{g'(s)}\right)\, ds$

And so i now come to the conclusion, that the adjoint operator is given by

$x\mapsto \int\limits_t^1 x(s)\, ds+x'(t)$.

Where is my mistake?? Or is it right now?

  • $\begingroup$ Why don't you add it to your question? It would be better visible. Just click "edit" under you question $\endgroup$ – Ilya Jan 11 '13 at 12:48
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    $\begingroup$ It is good now. In the version you wrote in you question statement you took the complex conjugate of the $x'$ term, which is incorrect. $\endgroup$ – Willie Wong Jan 11 '13 at 13:27
  • $\begingroup$ thank you very much for your help $\endgroup$ – math12 Jan 11 '13 at 13:29

Let $f$ and $g$ two test functions. We have $$\langle f,Tg\rangle_{H^1}=\int_0^1f(t)\overline{\int_0^tg(s)ds}dt+\int_0^1f'(t)\overline{g(t)}dt.$$ Integrating by parts, this gives \begin{align}\langle f,Tg\rangle_{H^1}&=\int_0^1f(s)ds\overline{\int_0^1g(t)dt}-\int_0^1\int_0^tf(s)ds\overline{g(t)}dt+\int_0^1f'(t)\overline{g(t)}dt\\ &=\int_0^1\int_t^1f(s)ds\overline{g(t)}dt+\int_0^1f'(t)\overline{g(t)}dt, \end{align} so $T^*f(x)=\int_x^1f(s)ds+f'(x)$.

  • $\begingroup$ Is my solution correct? I am confused now, sorry. My solution is that $T^*$ is given by $f\mapsto\int_t^1 f(s)\, ds + f'(t)$. You have $-f'(t)$. $\endgroup$ – math12 Jan 11 '13 at 21:12
  • $\begingroup$ Yes, it is (and I upvoted it). $\endgroup$ – Davide Giraudo Jan 11 '13 at 21:12
  • $\begingroup$ Thanks! You helped me very much with your answer. $\endgroup$ – math12 Jan 11 '13 at 21:30
  • $\begingroup$ @DavideGiraudo Could you elaborate a bit more on the differentiation by parts? $\endgroup$ – Brofessor Jun 12 '18 at 13:38

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