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Let $p,q\in(1,\infty)$ be such that $1/p+1/q=1$ and let $(\Omega, \mathcal A,\mu)$ be a $\sigma$-finite measure space.

Claim: The map $$\phi:L^q(\Omega)\to \left(L^p(\Omega) \right)^*,\quad \phi(g)(f)=\int_\Omega fgd\mu$$ is an isometric isomorphism.

Proving that $\phi$ is well-defined, linear and continuous was not too difficult. I also proved that $\|\phi(g)\|_{(L^p)^*}\leq \|g\|_{L^q}$ holds, but failed at showing the reverse inequality. This leads me to

Question 1: What would be a function $f\in L^p(\Omega)$ with $$\int_{\Omega} fgd\mu=\|g\|_{L^q}\quad ?$$

To prove that $\phi$ is an isomorphism, it suffices to prove that it is bijective. I can prove injectivity but not surjectivity, hence

Question 2: Why is $\phi$ surjective?

John Steinbeck
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  • you have to prove that $L^p(\Omega)^$ is exactly the set of functionnals $f \to \int_\Omega fg d\mu$ where $g \in L^q(\Omega)$. to achieve this, for a $G \in L^p(\Omega)^$ find a sequence $g_n \in L^q(\Omega)$ such that $\phi(g_n)$ converges to $G$ in $L^p(\Omega)^*$ and show that $g_n \in L^q(\Omega)$ converges to some $g$ in $L^q(\Omega)$ and hence that $G = \phi(g)$ – reuns Mar 22 '16 at 18:01
  • consider a $G \in L^p(\Omega)^$. for any $x_0 \in \Omega$ and $n \in \mathbb{N}$ let $u_{x_0,n}(x) = C_n e^{-n |x-x_0|^2 }$ with $\frac1{C_n} = \int_{\mathbb{R}^d} u_{x_0,n}(x) dx$. clearly $u_{x,n} \in L^p(\Omega)$ hence the functions $g_n(x) = G(u_{x,n})$ are well-defined. then it is easy to prove that $\phi(g_n)$ converges to $G$ in $L^p(\Omega)^$ and $|g_n|$ converges, hence $g_n$ converges to some $g$ in $L^q(\Omega)$, whence $G = \phi(g)$ – reuns Mar 22 '16 at 20:25

2 Answers2

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Question 1 : choose $f=g^{q-1}\cdot sign(f)$ Question 2 : Notice that $\phi \in \Big(L^q(\Omega) \Big)^{*}$ we ca define $$\nu (A)=\phi( \mathbb{1} _{A})$$, $\nu$ ia a measure and absolutly continuos acording to $\mu$ so find with Radon-Niodim Theorem we get $g$ that works for $f$ an indicator. We imediatly conclude that $g$ work for simple function. For $f$ general we have $f_n\to f$ simple functions that $|f_n|\leq|f|$ (the normal constraction) and so from the Dominated Convergence Theorem $g$ will work for all $f$. Q.E.D

Yonatan
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Let $f(x)=0$ when $g(x)=0.$ Let $f(x)=|g(x)|^{q-2}\cdot \overline {g(x)}$ when $g(x)\ne 0.$ Since $|f(x)|=|g(x)|^{q-1}$ and $p(q-1)=q,$ we have $|f(x)|^p=|g(x)|^q$ and $f(x) g(x)=|g(x)|^q.$

We have $\psi (g)(f)=\|g\|^q_q$ and $\|f\|_p=\|g\|_q^{q/p}=\|g\|_q^{q-1}$. So $$ \psi (g)(f)=\|g\|_q^q=\|g\|_q\cdot \|g\|_q^{q-1}=\|g\|_q\cdot \|f\|_p.$$ And $g\ne 0\implies f\ne 0.$

When $g\geq 0$ we can simply write $f=g^{q-1}.$

DanielWainfleet
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