The $L^p(\mathbb R)$ norm is increasing as a function of $p$ (Update: It's false!)

Question

Update: This is false. See the answers for a counterexample.

Let $C\ge 1$ be a constant. Fix $f\in L^p(\mathbb R)$ for $p\ge C$. Show that $$p\rightarrow \left( \int |f|^p \right)^{1/p}$$ is non-decreasing.

Comments: I'm posting this because there is (surprisingly) no good reference for this fact on the internet. If I recall correctly, differentiating with respect to $p$ will do the trick.

For the same problem on a finite measure space, see here.

Answer 1

Jensen:

take q>p, examine:

$L(q,p)$ = $\frac {(\int |f|^q)^{\frac 1q}}{(\int |f|^p)^{\frac 1p}}$

By Jensen's inequality $(\int |f|^p)^{\frac qp}$ ≤ $(\int |f|^q)$ Because $\frac qp$≥1 $\Rightarrow$ $g(x)$ = $x^{\frac qp}$ is convex

Thus, $L(q,p)^q$≥ $\frac {\int |f|^q}{\int |f|^q}$ = 1 $\Rightarrow$ $L(q,p)$≥1 $\Rightarrow$ $p \rightarrow (\int |f|^p)^{\frac 1p}$ is non decreasing

Answer 2

This is not true: Consider the function $f(x)=\chi_{\mathbb{R}\setminus(-1,1)}(x)|x|^{-1/2}$. Then for $p\leq 2$ we have $\| f\|_{p}=\infty$ and for $p>2$ we get $$ \| f\|_p = \left(\frac{4}{p-2}\right)^{1/p}. $$