Is this normed linear space a Banach space? 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?

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
