TITLE: Completitude of $L^1[0,1]\cap L^2[0,1]$ with the maximumm norm
QUESTION [7 upvotes]: I am trying to prove from scratch that the space $L^1[0,1]\cap L^2[0,1]$ equipped with the norm:
$$\left \| f \right \|=\max\{\left \| f \right \|_1,\left \| f \right \|_2\}$$
defines a Banach space, but I am having trouble checking completitude. I have considered the Cauchy sequence $\{f_n\}$, and found that by completitude of $L^1[0,1]$ and $L^2[0,1]$, there must be $f_0^1, f_0^2$ such that:
$$\left \| f_n -f_0^1\right \|_1\rightarrow 0$$
$$\left \| f_n -f_0^2\right \|_2\rightarrow 0$$
But I don't know how to prove that $\left \| f_0^1-f_0^2 \right \|=0$. Any ideas?

REPLY [2 votes]: Since $[0,1]$ has finite measure, thank's to Hölder inequality (Jensen works too) $L^2 [0,1] \subset L^1 [0,1]$ and :
$$\|f\|_1 \leq \lambda([0,1])^{\frac{1}{1} -\frac{1}{2} } \|f\|_2  = \|f\|_2$$
That's why your exotic space is the well-known $L^2[0,1]$ equipped with the well-known second norm, which is complete, of course :) .