Let $g$ be an $L^{2}$ function on $[0,2]$, with respect to the Lebesgue measure $m$. Is it true that $$||g||_{1}=\lim_{p\to 1}\left(\int |g|^{p}~dm\right)^{1/p}?$$
I'm really not sure how to tackle this problem. I'm. assuming that the integral is taken over $[0,2]$, but it wasn't explicitly given in the statement. I know that $\left(\int_{0}^{2}|g(x)|^{2}~dx\right)^{1/2}<\infty$ since $g\in L^{1}([0,2])$, so my intuition is that the questioned equality does hold. However, I'm not sure if this is correct, and if it is, I'm not sure how to prove it. Any help is appreciated.
Consider the map $$ h(p):=\int_{[0,2]}|g|^p. $$ It suffices to show that $h$ is continuous at $p=1$ since then the map $$ p\mapsto \|g\|_p=\exp\left(\frac{\log(h(p))}{p}\right) $$ is also continuous at $p=1$.
In order to show that $g$ is continuous at $p=1$, all you need is the dominated convergence theorem. But $L^2\supset L^p$ for all $0<p<2$ since you are in a space of finite measure. So you have a natural dominated function $|g|^2$ since $$ \|g\|_p\leq\|g\|_22^{1/p-1/2}. $$