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Problem: Let $f\in L^2(0,\infty)$ and let $(Tf)(s)=\frac1s\int_0^sf(t)dt$. Find the adjoint, $T^*$.

Attempt: I know that problems like these should be very simple, but oftentimes I find them very difficult, to my shame. I understand that the adjoint in this case is defined as being the $T^*:L^2(0,\infty)\to L^2(0,\infty)$ where, for all $f,g\in L^2(0,\infty)$,

$$\langle Tf,g\rangle=\langle f,T^*g\rangle$$

I understand the method one usually employs; namely to start from the left hand side and via manipulation (in this case, of the integrals) to get the inner product of $f$ with something else.

In this case though, I don't really understand what is going on with regards to the order of integration changing and how, exactly, the limits thereof change as well. I have been told it is to do with Fubini's theorem, but it's not something I have really grappled with, and had exposure to before.

I have had a look at some examples like this one here involving the same inner product, but I am thrown more than usual by the $1/s$ factor.

Could somebody use the above problem to elucidate what is going on, and how the machinations behind these manipulations really work?

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We have $$ \langle Tf,g \rangle= \int_0^\infty g(s)\left( \frac1s\int_0^sf(t)\mathrm dt\right)\mathrm ds\\ =\int_0^\infty\int_0^s\frac1sg(s) f(t)\;\mathrm dt\mathrm ds $$ we would like to break off $f$ in a sense, to get an integral of the form $\langle f,\cdot\rangle$, where the $\cdot$ will be the adjoint.

Here, the usual method is to use Fubini's, $$ \int_0^\infty\int_0^s\frac1sg(s) f(t)\;\mathrm dt\mathrm ds\\ =\int_0^\infty \int_t^\infty \frac1sg(s)f(t)\;\mathrm ds\mathrm dt\\ =\int_0^\infty f(t) \left(\int_t^\infty \frac1sg(s)\;\mathrm ds\right)\mathrm dt\\ $$ so $$ T^*g(t)=\int_t^\infty\frac{1}{s}g(s)\mathrm ds $$

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    $\begingroup$ (+1) A bit more detail about the motivation $\endgroup$ – robjohn May 5 '18 at 20:11
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With a change of the order of integration: $$ \int_0^\infty\overbrace{\frac1s\int_0^sf(t)\,\mathrm{d}t}^{Tf(s)}\,g(s)\,\mathrm{d}s =\int_0^\infty\overbrace{\int_t^\infty\frac1sg(s)\,\mathrm{d}s}^{T^\ast g(t)}\,f(t)\,\mathrm{d}t $$

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  • $\begingroup$ sorry, I was typing mine (slowly) when you posted yours. +1 also for the comment on the op $\endgroup$ – qbert May 5 '18 at 20:08

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