Do we have $\|x\|_p \le \|x\|_q$, where $x$ is bounded and $p>q$?

Question

Assume that $x\in L_p(0,\infty) \cap L_q(0,\infty)$, where $p>q$.

Do we have $$\|x\|_p \le \|x\|_q?$$ where $x$ is bounded.

Answer 1

The answer is no, $q < p$ does not generally imply $\|x\|_p \leq \|x\|_q$.

For a simple example, take $$x(t) = \begin{cases}1 & \text{ if }t \in [0,a] \\ 0 & \text{otherwise} \\ \end{cases}$$ Then $\|x\|_p = a^{1/p}$. If $a > 1$, then we have $a^{1/p} \downarrow 1$, so in this case indeed $\|x\|_p < \|x\|_q$. However, if $a < 1$, then $a^{1/p} \uparrow 1$, so the reverse inequality holds: $\|x\|_p > \|x\|_q$.

In the previous example, $p \mapsto \|x\|_p$ was monotone as a function of $p$. In general, this is not the case. For example, consider $$x(t) = \begin{cases}2 & \text{if }t \in [0,1/2] \\ 1 & \text{if }t \in (1/2, 5/2] \\ 0 & \text{otherwise} \\ \end{cases}$$ For this function, we have $\|x\|_p = (2^p(1/2) + 2)^{1/p}$. As you can see by plotting this as a function of $p$, the function is not monotonic: at $p=1$ its value is $3$, then it decreases to a minimum of approximately $1.8$ at around $p=4$, then it increases to approach $\|x\|_{\infty} = 2$ as $p$ grows large.

Note that we can add a hypothesis which will ensure monotonicity, although not in the direction you desire. If $x$ is a bounded function which is supported on a set $E$ with $\mu(E) \leq 1$, then we do have $\|x\|_p \uparrow \|x\|_{\infty}$ as $p \to \infty$, so $q < p$ implies $\|x\|_q \leq \|x\|_p$. This is straightforward to prove using Hölder's inequality: $$\|x\|_q = \|x\cdot \chi_E\|_q \leq \|x\|_p \|\chi_E\|_r = \|x\|_p \mu(E)^{1/r} \leq \|x\|_p$$ where $r$ is the number that satisfies $1/q = 1/p + 1/r$.