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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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3  
I don't think there is a general way to find an adjoint operator, but you can make a guess, then prove that it is actually what you want. The intuition I always resort to is thinking of an operator as a matrix. Its adjoint is then something similar to a conjugate transpose of the matrix. In your case, your operator is something like a lower triangular matrix (if you consider elements of $L^2([0,1])$ as column vectors). Its adjoint should be something like an upper triangular matrix: $(Sf)(x) = \int_x^1 f(s)dx$. This is just a guess. I have not verified. –  Tunococ Aug 14 '12 at 20:54
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The adjoint is given by $\langle Tf, g\rangle = \langle f, T^*g\rangle$, so a general idea could be writing down $\langle Tf, g\rangle$ and using integral manipulations to write it as the scalar product of $f$ with something. Then something is $T^*g$. –  martini Aug 14 '12 at 21:10
    
@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, –  Andrew Sep 7 '13 at 21:00

2 Answers 2

up vote 10 down vote accepted

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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sorry what did you do to get the last integral? –  anegligibleperson Aug 15 '12 at 0:23
    
@anegligibleperson: I just changed the order of integration. –  Mhenni Benghorbal Aug 15 '12 at 2:32

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". –  Fabian Aug 14 '12 at 21:37

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