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This is from my homework, I'm totally lost as to how to proceed. Consider the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $(Tf)(x) = \int^x_0 f(s) \ ds$ What is the adjoint of $T$?

This operator doesn't seem to be an orthogonal projection, nor is it self-adjoint. How does one find the adjoint of an operator in general? Thanks in advance!

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    @martini Thanks for this comment. I had been trying to reproduce some results I had seen and was stuck. This comment is most instructive. Regards,2013-09-07

2 Answers 2

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Using the fact that

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

we have

$ \langle Tf, g\rangle = \int_{0}^{1} (Tf)(t)g(t)\,dt =\int_{0}^{1} \int_{0}^{t} f(\tau)\,d \tau\, g(t)\, dt = \int_{0}^{1} f(\tau)\, \left(\int_{\tau}^{1} g(t) \,dt\right)\, d \tau $ $ = \langle f, T^{*}g\rangle $

From the last integral, we can see that the adjoint is given by

$ (T^{*}f) (x) = \int_{x}^{1} f(s)\, ds $

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    Nevermind, I guess [the comment to the accepted answer in this duplicate post](https://math.stackexchange.com/a/242340/405143) explains that; saying it is just the triangle $\{(s,t) \mid 0 \le s \le t \le 1\}$.2018-03-23
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We can find adjoint for kernel operators, that is, operators given by $T(f)(x)=\int_{[0,1]}K(x,y)f(y)dy,$ with $K$ satisfying good conditions. We should have $\int_{[0,1]}T^*(f)(x)\overline{g(x)}dx=\int_{[0,1]}f(x)\overline{T(g)(x)}dx.$ Since $\int_{[0,1]^2}f(x)\overline{K(x,y)g(y)}dxdy=\int_{[0,1]}\left(\int_{[0,1]}\widetilde K(y,x)f(x)dx\right)\overline{g(y)}dy,$ where $\widetilde K(x,y)=\overline{K(y,x)}$. Since it's true for any $g$, we have $T^*(f)(x)=\int_{[0,1]}\widetilde K(x,y)f(y)dy.$

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    Now the relation to Tunococ's comment above is clear. The adjoint kernel is just the "conjugate-transpose".2012-08-14