Completitude of $L^1[0,1]\cap L^2[0,1]$ with the maximumm norm 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?
 A: You can use a lemma, which regards $L^p$:

If a sequence $f_1,f_2,\ldots$ converges to $f$ in $L^p$, then some subsequence $f_{s_1},f_{s_2},\ldots$ converges pointwise almost everywhere to $f$.

The idea to prove this is to pick some subsequence such that
$$\|f_{s_i}-f\|<2^{-i}$$
You can then note that the set of points $x$ that don't converge pointwise to $f$ in this sequence is those such that there exists some $\varepsilon > 0$ such that for all $i$ we have $|f_{s_i}(x) - f(x)| >\varepsilon$. For each $\varepsilon$ and $i$, you can let $E_{\varepsilon,i}$ be the set of $x$ such that  $|f_{s_i}(x) - f(x)| >\varepsilon$. This set must have measure less than $\frac{2^{-i}}{\varepsilon^p}$.
Then, you can define $E_{\varepsilon}=\bigcap_{i=1}^{\infty}\bigcup_{j=i}^{\infty}E_{\varepsilon,j}$ to be the set of points that fail to converge by a specific $\varepsilon$. The inner union has measure at most $\frac{2^{1-i}}{\varepsilon^p}$ by summing up the measures of the unioned sets - but $E_{\varepsilon}$ is an intersection of all of these over all $i$, so must have measure $0$ since it is an intersection of sets whose measure tends to $0$.
Then you may define $E=\bigcup_{n=1}^{\infty}E_{1/n}$ to be the set of points that fail to converge, but this is a union of measure zero sets, so has measure zero itself. Therefore $f_{s_i}$ converges pointwise almost everywhere to $f$.
It's very likely that an argument almost identical to this one appeared in the proof that $L^p$ was complete - an argument very similar to this one works to show that if the sequence was Cauchy, then it converges pointwise almost everywhere to something. (There are other ways to show $L^p$ is complete, but they are less common).
With this lemma, you can show that $f_0^1$ and $f_0^2$ are equal almost eveywhere: Since $f_i$ converges to $f_0^1$ in $L^1$, pick a subsequence of it converging pointwise almost everywhere to $f_0^1$. This subsequence still converges to $f_0^2$ in $L^2$. Pick a subsubsequence of this converging pointwise almost everywhere to $f_0^2$. Now you have a sequence which converges pointwise almost everywhere to both $f_0^1$ and $f_0^2$, so they must be equal almost everywhere.
A: 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 :) .
