I could not understand the last part of the proof of the following theorem: Let $ p\geq 1 $ and $ g $ be a measurable function such that $ \int|fg|dm<\infty $ for every $ f\in L^{p}(\mathbf{R}) $. Prove that there is a constant $ M $ such that $ \int|fg|dm\leq M \| f\|_{L^{p}} $ for all $ f\in L^{p}(\mathbf{R}) $.
Proof. Let $$ g_{n}(x)=\begin{cases}g(x)&\text{if $|g(x)|\leq n$ and $|x|\leq n$},\\ 0&\text{otherwise}\end{cases}$$ and $ \frac{1}{p}+\frac{1}{q}=1 $. Then $ g_{n}\in L^{q} $ for all $ n $. We observe that $ |g_{1}f|\leq |g_{2}f|\leq.........\leq |gf| $. So $ \lim_{n\rightarrow \infty} g_{n}(x)f(x)=f(x)g(x) $ for all $x$. It follows that the sequence of bounded linear functionals $ f\mapsto \int f(x)g_{n}(x)dx $ on $ L^{p}(\mathbf{R}) $ ( since $ g_{n}\in L^{q} $, this linear functionals are bounded) converges to $ f\mapsto \int f(x)g(x)dx $. Then the Banach- Steinhaus Theorem, $ f\mapsto \int f(x)g(x)dx $ is bounded linear functional on $ L^{p}(\mathbf{R}) $. So $ f\mapsto \int f(x)g(x)dx $ is continous on $ L^{p}(\mathbf{R}) $ and therefore there is an $ h\in L^{q}(\mathbf{R}) $ such that $\int f(x)g(x)dx=\int f(x)h(x)dx $, which then implies that $ g=h $ a.e.
I could not understand why $ h $ exists. I think $ h $ should exist because of the continuity of linear function. But I could not understand the relation.
You have proved that the map $f\mapsto \int f(x)g(x)\,\mathrm dx$ is a continuous linear functional on $L^{p}(\mathbb{R})$. Now use the isomorphism between the dual of $L^{p}(\mathbb{R})$ and $L^{q}(\mathbb{R})$ to find the function $h\in L^{q}(\mathbb{R})$.