Is this normed linear space a Banach space?

Question

Let $E$ be a measurable set of finite measure and $1 < p_1 < p_2 < \infty$. Consider the linear space $L^{p_2} (E)$ normed by $||.||_{p_1}$ . Is this normed linear space a Banach space?

Answer 1

I left about three wrong comments due to mixing up $p_1$ and $p_2$, so let me say it the right way this time.

It is not a Banach space, in general. Take $E = (0,1)$ with Lebesgue measure, for instance, and $p_1 = 2$, $p_2 = 4$. The function $f(x) = x^{-1/3}$ is in $L^2$ but not in $L^4$. Let $f_n$ be a sequence in $L^4$ which converges, in $L^2$ norm, to $f$; for instance, $f_n = \max(f, n)$. This sequence is Cauchy in $L^2$ norm, but it does not converge in $L^2$ norm to any element of $L^4$.

One can show it is only a Banach space if $E$ is a finite union of atoms. In such a case $L^{p}(E)$ is finite dimensional for all $p$ and so all norms are equivalent.

For practical purposes, you should expect that a given vector space has at most one "reasonable" Banach norm, up to equivalence. A second norm on a Banach space is either equivalent to the first one, or is incomplete, or is some pathological non-constructive axiom of choice monster. You can read more about this in Schechter, Handbook of Analysis and its Foundations. In this case the norms are clearly not equivalent (unless, as mentioned, $L^p(E)$ is finite dimensional), and they are both very explicit, so we expect they can't both be complete.

Answer 2

Okay lets suppose that $(L^{p_2},\|.\|_{p_1})$ is a Banach space then the map \begin{align*} \Phi:(L^{p_2},\|.\|_{p_2}) &\to (L^{p_2},\|.\|_{p_1}), \\f &\mapsto f \end{align*} is not only continuous but also bicontinuous. This follows from the fact that $\|.\|_{p_1} \leq C \|.\|_{p_2}$ and the open map theorem. Now this yields that there is a $K>0$ such that \begin{align*} \|\Phi(f)\|_{p_1} \geq K \|f\|_{p_2} \\ \|f\|_{p_1}\geq K \|f\|_{p_2} . \end{align*} Now take a sequence of sets $A_n \subseteq E$ with $0<\lambda(A_n)\to 0$ ($\lambda$ denotes the Lebesgue measure) and consider the previous inequality \begin{align*} \lambda(A_n)^{\frac{1}{p_1}} \geq K \lambda(A_n)^{\frac{1}{p_2}} \quad \text{for all} \quad n\in\mathbb{N}\\ \lambda(A_n)^{\frac{1}{p_1}-\frac{1}{p_2}} \geq K \quad \text{for all} \quad n\in\mathbb{N} \end{align*} But $\frac{1}{p_1}-\frac{1}{p_2}>0$ because of our assumptions, this leads to $K = 0$ which contradicts $K>0$. So it can't be Banach space