In Stein & Shakarchi's Functional analysis book, the following lemma is used for the proof of Riesz Thorin interpolation theorem.
Let $p\in[1,\infty]$ and $p'$ the conjugate of $p$, i.e. so that $1/p+1/p'=1$. Then for $f$ that is integrable on any set of finitemeasure $\|f\|_p=\sup_{g\text{ simple with }\|g\|_{p'}=1}\left|\int fg\right|$.
Correct me if I am wrong, it seems like the book only proved the lemma for real-valued functions, however the Riesz Thorin interpolation theorem requires the underlying field to be complex, so we need the lemma to work for complex-valued function. However, the proof for the real version make use of the sign function and it being a simple function, but the complex analog $f/|f|$ is not simple. So I don't know how to generalised the proof to the complex case.
After a lot of effort, I think I have come up with a proof, but it doesn't work for the case $p=1$
If $\|f\|_p<\infty$ and $p\not=1$, then by the Riesz representation theorem, $L^p\cong(L^{p'})'$ isometrically via $f\mapsto\langle\cdot,f\rangle$, we have $$\|f\|_p=\sup_{\|g\|_{p'}=1}\left|\int fg\,d\mu\right|$$ We can pick $g_n$ simple with $g_n\to g$ and $|g_n|\leq|g|$, then $\|g_n\|_{p'}\leq 1$ and by dominated convergence $\int fg_n\,d\mu\to\int fg\,d\mu$. Hence the result. If $p=1$, we can just take $g=\bar f/|f|$.
Suppose now $\|f\|_p=\infty$ and $p\not=1$, we need to show that the RHS of $(*)$ is not finite. It suffice to show that if the RHS of $(*)$ is finite, then $\|f\|_p<\infty$. Denote the RHS of $(*)$ as $M$. Let $D$ be the space of all simple function that belongs to $L^{p'}$. Define $T:D\to\mathbb C$ by $T(g)=\int fg\,d\mu$, this integral indeed exist since $f$ is locally integrable and $g$ is simple. Now $T$ is a bounded linear map with norm $M$. Since $D$ is dense in $L^{p'}$, the map $T$ can be extended uniquely to a bounded linear map $L^{p'}\to\mathbb C$ with norm $M$. By the Riesz representation theorem, $T$ is represented by some $\tilde{f}\in L^p$. We know $$\int\underbrace{\Re(f-\tilde{f})}_{:=h}\mathbf 1_{A}d\mu+i\int\Im(f-\tilde{f})\mathbf 1_{A}d\mu=\int(f-\tilde{f})\mathbf 1_{A}d\mu=0$$ for any set $A$ of finite measure. So $\int h\mathbf 1_A\,d\mu=0$ for any set $A$ of finite measure. We claim that $h=0$ a.e. Suppose not, then there exist $\epsilon>0$ and a set $A$ of finite non-zero measure such that $\epsilon\mathbf 1_A\leq|h|$. Then $$0<\epsilon\mu(A)\leq\int|h|d\mu=\int h(\mathbf 1_{A\cap\{|h|>0\}}-\mathbf 1_{A\cap\{|h|<0\}})d\mu=0.$$ This is a contradiction, so $h=\Re(f-\tilde{f})=0$ a.e. Similarly, $\Im(f-\tilde{f})=0$ a.e. So $f=\tilde{f}$ a.e. So $\|f\|_p=M$.
Is this proof for the $p\not=1$ case correct? Is there a way to prove the $p=1$ case? Is there a direct way to generlised the proof found in Stein & Shakarchi's book without appealing to the Riesz representation theorem? (The book actaully used this lemma to prove the Riesz representation theorem) Thanks very much in advance!