# For which $p, q$ exists $C > 0$ with $||f||_p \leq C ||f||_q$ for all $f \in C([0,1])$?

I want to find out for which $p, q$ exists $C > 0$ with $||f||_p \leq C ||f||_q$ for all $f \in C([0,1]), p \in \mathbb{R}_{\geq 1} \cup \{\infty\}$. I first let $p,q \geq 1$ and I looked at the case where $q > p$.

I defined $\tilde{f} := |f|^p$ and $\tilde{p} := \frac{q}{p}$, $f \in C([0,1])$ of course. Also, I let $g(x) = 1$ and applied Hölder's inequality:

$$||\tilde{f} \cdot g||_1 \leq ||\tilde{f}||_\tilde{p} \cdot ||g||_\tilde{q}$$

where of course $||g||_\tilde{q} = 1$.

$$\implies \int_0^1 |f|^p \mathrm dx \leq \left ( \int_0^1 |f|^q \mathrm dx \right )^\frac{p}{q} \iff ||f||_p \leq ||f||_q \cdot C \;\; \forall C \geq 1$$

So my solution is, that for $q > p \geq 1$, there always exists $C > 0$ such that the inequality holds good. Now, before I look at $p > q$, I wanted to know whether my solution is correct. It doesn't seem to be complete to me, as there might also exist $C \in (0,1)$ such that the inequality holds good. Am I wrong? If not, how can I find out for which $p$ and $q$ there exists a $C \in (0,1)$?

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You have proved what you set out to: if $q > p \ge 1$ there exists $C > 0$ (namely, $C=1$) such that $||f||_p \le C ||f||_q$ for all $f \in C([0,1])$.

Note you are not trying to prove that $||f||_p \le C ||f||_q$ for all $C > 0$. This is easily seen to be false.

You might ask whether this result is sharp; whether there exists a number $0 < C < 1$ which could work. To find out, consider $f=1$.

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Thanks a lot for this answer. I indeed haven't considered the case $f=1$. – Huy Mar 13 '11 at 16:48

Your argument seems to be correct. It shows that indeed $\Vert f\Vert_{p} \leq \Vert f\Vert_{q}$ for $1 \leq p \leq q \lt \infty$.

Instead of using Hölder's inequality, you could also argue with Jensen's inequality (which is often used in the proofs of Hölder's inequality) as follows: For $p \leq q$ the function $\varphi(x) = x^{q/p}$ is convex, hence $\left(\int |f|^p \right)^{q/p} = \varphi\left(\int |f|^p\right) \leq \int \varphi(|f|^p) = \int |f|^{q}$ and extracting $q$th roots on both sides gives the desired $\Vert f \Vert_{p} \leq \Vert f \Vert_{q}$ for $p \leq q \lt \infty$. The case $q = \infty$ needs to be dealt with separately in this argument and I leave this to you.

To see that $C = 1$ is the best possible constant simply take $f$ to be the constant function $f = 1$, which has $\Vert f \Vert_{p} = 1$ for all $p$, so there can't be a constant $C \lt 1$.

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You say my argument shows the inequality for $1 \leq p \leq q < \infty$. But in my question, I also allow $p = \infty$. What exactly changes in that case? – Huy Mar 13 '11 at 17:24
@Huy: The case $q = \infty$ does not work as written because you'd integrate $|f|^{\infty}$. The easiest way to remedy this is to replace $f$ by $g = f/\Vert f\Vert_{\infty}$ (assuming $f \neq 0$ of course). Then observe that $\Vert g\Vert_{p} \leq 1$ and use that the $p$-norm is homogeneous, so $\Vert f\Vert_{p} \leq \Vert f\Vert_{\infty}$ by multiplying the last inequality by $\Vert f\Vert_{\infty}$. – t.b. Mar 13 '11 at 17:39