For $1 \leq r < p < \infty$ prove the continuous injection of $L^p([0, 1])$ into $L^r([0, 1])$.

Question

For $1 \leq r < p < \infty$ prove the continuous injection of $L^p([0, 1])$ into $L^r([0, 1])$.

I am having a hard time starting. Any suggestions. I tried a straight forward approach. That is, given $\epsilon > 0$, I tried to find a $\delta >0$ such that $||f - g||_p < \delta$ implies that $||f - g||_r < \epsilon.$

Thanks for any help.

Answer 1

Alternatively you can use Hölder's Inequality to find that $$ \lVert x \rVert_r \le \lVert x \rVert_p \tag{1} $$ and then you can proceed by letting $\epsilon > 0$ and putting $\delta = \epsilon$ so that $$ \lVert x - y \rVert_p < \delta \implies \lVert i(x) - i(y) \rVert_p < \epsilon \implies \lVert i(x) - i(y) \rVert_r < \epsilon $$

Hint for (1): $$ \frac{1}{p/r} + \frac{1}{p/(p-r)} = 1 $$

Answer 2

We have $\|x\|_r \le \|x\|_p$ (from Jensen's inequality, using $\lambda[0,1] = 1$).

Let $i:L^p[0,1] \to L^r[0,1]$ be the injection. Then $\|i(x)-i(y)\|_r = \|x-y\|_r\le \|x-y\|_p$, hence it is Lipschitz continuous.

This nesting is true more broadly for $L^p(X,\mu)$, where $\mu X < \infty$ (except the Lipschitz constant changes).