# Embeddings of norms

• Mar 16th 2009, 03:37 PM
HTale
Embeddings of norms
Just in case, here is a definition that I'm working with

Let $X$ be a linear space, with the norm $||\cdot||_X$, and $Y$ be a linear space with the norm $||\cdot||_Y$. If $X$ is a subspace of $Y$ and there exists a constant $C$ such that for all $f \in X$, one has $||f||_Y \leq C||f||_X$, then we write $X \subset Y$, and we say that there is an embedding of $X$ into $Y$.

Now, I want to prove that if $p < r$, then $l^p \subset l^{\infty}$. The hint to this question says the following

"First prove that $l^p \subset l^{\infty}$. Then write $r=p+s, s>0$ and use the inequality $|x_i|^r \leq |x_i|^p||x||_{\infty}^s$"

I don't think I've understood the hint properly, because somehow I'm not using the first half of the hint. I've found that the second part of the hint is sufficient, here's what I've done so far;

We know that $|x_i|^s \leq (\sup_i |x_i|)^s = ||x_i||_{\infty}^s$. Now, we know that $r=s+p$, and so we have $|x_i|^r \leq |x_i|^p||x||_{\infty}^s$. Then,

$\sum_i |x_i|^r \leq \sum_i |x_i|^p||x||_{\infty}^s = ||x||_{\infty}^s\sum_i |x_i|^p$. Now, given that $r > p$, then $\frac{1}{r} < \frac{1}{p}$,

and so

$\left(\sum_i |x_i|^r\right)^{1/r} \leq ||x||_{\infty}^{s/p}\left(\sum_i |x_i|^p\right)^{1/p}$

which is just $||x||_r \leq ||x||_{\infty}^{s/p}||x||_p$. Comparing this to the definition, we find that $l^p \subset l^r$.

Have I missed something here, I'm not entirely sure if this is correct.