Norms are not equivalent in infinite spaces whenever they lie in $L_1$ space

Question

This is a problem from Steward's book:

Let $L_1$ be a set of infinite sequences $x_i$ with $\sum\limits_{j=1}^{\infty} |x_j^{(i)}|<\infty$. Show that if $1\leq m<n$, then there is a sequence $\{x_k\}$ in $L_1$ such that $||x_k||_m\to \infty$, but $||x_k||_n\to 0$.

I have several failed tries, e.g. I set $x_k=(\frac{1}{k})^{\frac{1}{m}+1}$. However, it does not work at all: It converges in both norms.

I am wondering how to find such a sequence that diverges in $m$_norm but converges in $n$_norm. There should be a trick?

Answer 1

For simplicity, let $n=2$ and $m =1$.

Now, let us consider the sequence \begin{align} x^{(i)}_j = \frac{1}{j^{1/2}i^{1+1/j}} \end{align} then we see that \begin{align} \|x_j\|_2^2 = \sum^\infty_{i=1} \frac{1}{ji^{2+2/j}}\leq \sum^\infty_{i=1}\frac{1}{ji^2}\leq\frac{C}{j} \end{align} but \begin{align} \|x_j\|_1 = \sum^\infty_{i=1} \frac{1}{j^{1/2}i^{1+1/j}} \geq \sqrt{j} \rightarrow \infty \end{align} as $j\rightarrow \infty$.

For an arbitrary pair $1 \leq m < n$, the construction is more or less the same with a slight change to the powers of the above example.