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I see that the above operator is bounded.

I ended up with an argument to calculate the adjoint as follows,

$$ <f,Bg>=\int_0^1\overline{f(x)} \int_0^xg(t)\,dt\,dx $$

I see $f(x)$ as the derivative of $\frac{d}{dt}\int_0^t f(x)\,dx$ and use integration by parts in above equation and get

$$ <f,Bg>=-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\big[\int_0^xg(t)\,dt \,\int_0^x\overline{f(t)}\,dt\big]_0^1 $$

So domain of $B^*$ is $f\in L^2(0,1)$ such that $\int_0^1\overline{f(t)}\,dt=0$

But B is a bounded operator, domain of adjoint should be whole space $L^2(0,1)$.

I do not know is the mistake?

Please help

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  • $\begingroup$ What does Fubini's theorem give you, changing order of integration instead of integrating by parts? $\endgroup$
    – mickep
    Mar 22 '15 at 19:51
  • $\begingroup$ @mickep I am sorry as the domain in second integral is dynamic, I do not know how fibula theorem applies. Please clarify. $\endgroup$
    – Harish
    Mar 22 '15 at 19:56
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    $\begingroup$ $\langle f,Bg\rangle=\int_0^1 \overline{f(x)}\int_0^x g(t)\,dt \,dx = \iint_D \overline{f(x)}g(t)\,dx\,dt = \int_0^1 \int_t^1 \overline{f(x)}\,dx g(t)\,dt = \langle Sf,g\rangle$, where $D=\{(t,x)~|~0<t<x,\ 0<x<1\}=\{(t,x)~|~t<x<1,\ 0<t<1\}$ and $(Sf)(t)=\int_t^1 f(x)\,dx$. I hope this became correct, without preview... $\endgroup$
    – mickep
    Mar 22 '15 at 20:00
  • $\begingroup$ @mickep Thanks, if you could tell me the mistake in my arguments. Thanks for the explanation. $\endgroup$
    – Harish
    Mar 22 '15 at 20:09
  • $\begingroup$ You will essentially get the same thing from the integration by parts, since $\int_0^1-\int_0^x = \int_x^1$. $\endgroup$
    – mickep
    Mar 22 '15 at 20:10
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After integrating by parts, you get $$ \langle f,Bg\rangle =-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\big[\int_0^xg(t)\,dt \,\int_0^x\overline{f(t)}\,dt\big]_0^1 $$ Continuing this calculation (inserting the limits, noting that $x=0$ gives nothing, and renaming the integrating variable in the $g$ integral), we get $$ \begin{aligned} \langle f,Bg\rangle &=-\int_0^1\int_0^x\overline{f(t)}g(x)\,dt\,dx +\int_0^1g(x)\,dx \,\int_0^1\overline{f(t)}\,dt\\ &=\int_0^1g(x)\biggl[\int_0^1\overline{f(t)}\,dt-\int_0^x \overline{f(t)}\,dt\biggr]\,dx\\ &=\int_0^1g(x)\int_x^1\overline{f(t)}\,dt\,dx\\ &=\langle Sf,g\rangle, \end{aligned} $$ where $$ (Sf)(x)=\int_x^1 f(t)\,dt. $$

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  • $\begingroup$ Thank you so much for your answer, now I see I missed the integral in "non boundary term". @mickep $\endgroup$
    – Harish
    Mar 22 '15 at 20:33

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