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.

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

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 $T^*\,:\,(L^q)^*\rightarrow(L^p)^*$.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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