1
$\begingroup$

Let $E = L^p(0,1)$ with $1 ≤ p < ∞$. Given $u ∈ E$, set

$$Tu(x):=\int_0^x u(t)dt$$

Find the adjoint of $T$.

I know how to this in the case $p=2$ as shown here. But in general $L^p$ is not an Hilbert space and the definition of adjoint is different: $$\langle Tx,y^*\rangle =\langle x,T^*y^*\rangle $$ with $T:X\to Y$, $y^*\in Y^*$ and $T^*:Y^*\to X^*$.

In this case I do not necessarily have an integral representation of the duality and therefore I do not know how to compute the adjoint.

$\endgroup$
  • $\begingroup$ Try to understand my solution and then give this problem a try. $\endgroup$ – Mhenni Benghorbal Jan 21 '15 at 11:13
3
$\begingroup$

As it turns out, you do still have integral representation of the duality. If $1\leq p<\infty$, the dual of $L^p$ is isometrically isomorphic to $L^{p'}$ where $p'$ is the dual exponent, i.e. $\frac1p+\frac1{p'}=1$. (The dual exponent of $1$ is $\infty$.) The duality is given by $$\langle f,g\rangle=\int_0^1f(x)g(x)dx$$ for $f\in L^{p'}(0,1),g\in L^p(0,1)$. One can show that $T\in\mathcal{L}(L^p,L^q)$ for all $1\leq p,q\leq\infty$, so assuming $p,q<\infty$ we have for all $f\in L^{q'},g\in L^p$, $$\langle f,Tg\rangle_{L^q}=\int_0^1\int_0^xf(x)g(y)dydx=\int_0^1\int_y^1f(x)g(y)dxdy=\langle T^*f,g\rangle_{L^p}$$ if we define $T^*f:=\int_y^1f(x)dx$. This defines the adjoint operator $T^*\,:\,(L^q)^*\rightarrow(L^p)^*$.

$\endgroup$
  • $\begingroup$ when you say "the duality given by", what is the reason behind it? I mean, how can I be sure that the duality is exactly given by that expression and not another one? $\endgroup$ – mastro Jan 21 '15 at 11:42
  • 1
    $\begingroup$ Sorry, that should be "the duality IS given by", I've edited that in. What I mean is that if $\Phi\in\mathcal{L}(L^{p'},(L^p)^*)$ is the canonical isometric isomorphism, then for all $f\in L^{p'},g\in L^p$ $$\langle \phi f,g\rangle=\int fg.$$ We usually identify $L^{p'}$ and $(L^p)^*$ rather than worry about $\Phi$. $\endgroup$ – Jason Jan 21 '15 at 12:00
  • $\begingroup$ Thank you very much. My background is in mathematical engineering BSc and just now I am doing functional analysis for my MSc in applied math so I lack knowledge in pure formal math and therefore, even though I can intuitively see what you mean, I do not really grasp it fully. In particular I did not know that $\phi$ being the canonical isometric isomorphism implies that we can express the duality as an integral. Actually I do not know what a canonical isometric isomorphism is. Thank you very much for the detailed answer!! $\endgroup$ – mastro Jan 21 '15 at 12:15
  • 1
    $\begingroup$ It's probably a good exercise for you to try! Define $\Phi(f)$ by $$\langle\Phi(f),g\rangle_{L^p}:=\int fg$$ for $f\in L^{p'},g\in L^p$ (the underlying measure space is not important here). You need to show a few things: (1) $\Phi(f)\in(L^p)^*$, i.e. $\Phi(f)$ is a bounded linear map (use Holder's inequality), (2) $\Phi$ itself is linear (3) $\Phi$ is an isometry, i.e. $\|\Phi(f)\|=\|f\|_{L^{p'}}$ and (4) $\Phi$ is surjective, i.e. for every $g^*\in(L^p)^*$ there exists $f\in L^{p'}$ such that $g^*=\Phi(f)$. (Injectivity follows from the isometry property.) $\endgroup$ – Jason Jan 21 '15 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.