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I am wanting to disprove (show that it is not the case) that for a sequence ${x_n} = x^n$, ($n\in\mathbb N$), that if $(x_i)^n \to x_i$ in $\mathbb R$,then $x^n\to x$ in $\ell_2$. I have gotten as far as using the $\ell_2$ norm to attempt to show that it doesn't converge, but I am somehow stuck. I have gone ahead and stated that if we get that $(x_i)^n \to 0$ as n approaches infinity, then $\sqrt{({\sum_{n}^{} ({x^n-0})^2})}$, but I am not getting much after this. Any help is greatly appreciated, thanks

Edit: ${x_n}$ is a sequence in $l_2$ equal to $x^n$, which is the n-th power. The index $i$ ranges from $i=1,2,3,...$. Thus, $(x_i)^n$ denotes the n-th power of the i-th coordinate.

Vegeta the Prince of Saiyans
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As A.B pointed out, my previous answer was wrong. In fact, the opposite is true!

I will use this fact: for every $x\in \ell_\infty$, $y\in \ell_2$, $$ \lVert x y \rVert_2 \le \lVert x \rVert_\infty \lVert y \rVert_2 $$ where $xy$ is point-wise multiplication.

In our case, we have a sequence $x = (x_i)_{i=1}^\infty \in \ell_2$, but are are also assuming that $\lim_{n\to \infty} x_i^n =0$. This implies that for each $i$, $\lvert x_i \rvert < 1$, or in other words that $\lVert x \rVert_\infty < 1$.

Then we have that $$ \lVert x^n \rVert_2 \le \lVert x \rVert_\infty^{n-1} \lVert x \rVert_2 \to 0 \quad \text{as } n\to\infty $$

The old (wrong) answer for reference:

Point-wise convergence in $l_2$ (which is what I understood you are talking about) is a weaker condition than convergence in norm. Since convergence in norm implies point-wise convergence, the best way to answer your question is to find a counter-example (or show that it exists), since what you are trying to prove can be true in some particular cases.

Here is such an example: Let $x_i = 2^{-i}$ for $i=1,2,\dots$ . Then $\Vert x \Vert^2 = \sum_i 2^{-2i} < \infty$, so $x \in l_2$.

Since $\lim_n x_i^n = 0$, we have that $x^n$ converges point-wise to $0$. But $$ \Vert x^n \Vert = \left ( \sum_{i=1}^\infty 2^{-2in} \right)^{1/2} = \left( \frac{4^n}{4^n - 1} \right)^{1/2} \to 1 $$ and therefore $x^n$ can not converge in norm to $0$.

Angelo Lucia
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