Showing that $\vert \vert x \vert \vert _{p} \xrightarrow{ p \to \infty} \vert \vert x \vert\vert_{\infty}$ for $x \in \ell^{1}$

Question

Show that $\vert \vert x \vert \vert _{p} \xrightarrow{ p \to \infty} \vert \vert x \vert\vert_{\infty}$ for $x \in \ell^{1}$

Let $0\neq x \in \ell^{1}$ and $p\in ]1,\infty[$ then $\vert \vert x \vert \vert _{p}=\vert\vert x ^{p-1}x\vert\vert_{1}^{\frac{1}{p}}\leq (\vert\vert x^{p-1}\vert\vert_{\infty}\vert\vert x\vert\vert_{1})^{\frac{1}{p}}$

now the term $\vert\vert x\vert\vert_{1}^{\frac{1}{p}}$ is negligible since $ x\in \ell^{1}$. So looking at $\vert\vert x^{p-1}\vert\vert_{\infty}^{\frac{1}{p}}$. I am unsure whether I can say $\vert\vert x^{p-1}\vert\vert_{\infty}^{\frac{1}{p}}=\vert\vert x \vert\vert^{\frac{p-1}{p}}_{\infty}$. If I can, then it's simply a case of taking the limit $p \to \infty$ so that we get the inequality $\lim\limits_{p\to \infty}\vert\vert x \vert\vert_{p}\leq \vert \vert x \vert \vert_{\infty}$

and since for any $p \in [1, \infty[: \vert \vert x \vert \vert_{\infty}\leq \vert \vert x \vert \vert_{p}$ we are done.

Answer 1

You have almost got the proof. However, your notations are bad. You should write $\sum|x_n|^{p} \leq (\sup_n|x_n|^{p-1}) \sum |x_n|$. It is true that $\sup_n|x_n|^{p-1} =(\sup_n|x_n|)^{p-1}$. Rest of the argument is OK.

Answer 2

That follows directly from the definition and the monotonicity of $t\mapsto |t|^{p-1}$ for nonnegative $t$: $$ \|x^{p-1}\|_\infty = \sup_k |x_k^{p-1}|= \sup_k |x_k|^{p-1} = (\sup_k |x_k|)^{p-1} = \|x\|_\infty^{p-1}. $$