Finding the adjoint of an operator

Question

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!

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
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

Mhenni Benghorbal · Accepted Answer · 2013-09-08T00:54:21.280

39

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

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

edited Sep 08 '13 at 00:54

answered Aug 14 '12 at 23:41

Mhenni Benghorbal

48,099

sorry what did you do to get the last integral? – anegligibleperson Aug 15 '12 at 00:23
@anegligibleperson: I just changed the order of integration. – Mhenni Benghorbal Aug 15 '12 at 02:32
@MhenniBenghorbal how would you find the adjoint if $T:L^p(0,1)\to L^p(0,1)$ ? In this case the definition of adjoint is different because $L^p$ is not an Hilbert space in general. – mastro Jan 21 '15 at 10:21
1

@anegligibleperson Note that changing the order of integration here is valid because of Fubini's theorem. – Math1000 Jun 18 '16 at 09:40
Could you elaborate on that last step ? I don't doubt that Fubini applies, but I simply don't see how the integrals change from $\int_{[0,1]} \int_{[0,t]} \dots ds dt$ to $\int_{[0,1]} \int_{[s,1]} \dots dt ds$. Also, could you verify that the fact that you use is the Hermitian adjoint ? – BadAtLaTeX Mar 23 '18 at 13:50
Nevermind, I guess the comment to the accepted answer in this duplicate post explains that; saying it is just the triangle ${(s,t) \mid 0 \le s \le t \le 1}$. – BadAtLaTeX Mar 23 '18 at 14:15
@MhenniBenghorbal the inproduct on the square integrable functions has defined with the complex conjugate in my notes, why can that be omitted here? – Mee98 Jan 09 '20 at 13:50
@Mee98 No. it should not be omitted here. They assume the functions are real. You should use complex conjugate – JacobsonRadical Nov 19 '20 at 14:34

Davide Giraudo · Answer 2 · 2012-08-14T21:35:23.793

21

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 $$T^*(f)(x)=\int_{[0,1]}\widetilde K(x,y)f(y)dy.$$

edited Aug 14 '12 at 21:35

answered Aug 14 '12 at 21:23

Davide Giraudo

181,608

6

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

Finding the adjoint of an operator

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