Duality of $L^p$ spaces

Question

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?

Answer 1

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

Answer 2

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 $f=g^{q-1}.$