# $p\mapsto\Vert f\Vert_{L^p}$ is continuous. [duplicate]

Could someone help me prove the following:

Let $(\Omega,\mathscr{A},\mu)$ be a probability space. Let $f:\Omega\rightarrow\mathbb{R}$ be a non-negative measurable function. How do I prove that the funtion $N_f:p\in [1,\infty)\mapsto \Vert f\Vert_p\in[0,\infty]$ is continuous?

I already know that this function is non-decreasing. So far, I've tried the following:

Let $p\geq 1$. First suppose that $\Vert f\Vert_p<\infty$. Let $r_n<p$ be a sequence with $r_n\rightarrow p$. Since $|f|^{r_n}<\max (1, |f|^p)$, which is in $L^1(\Omega)$, and $|f|^{r_n}\rightarrow |f|^p$, then by dominated convergence, we get $\Vert f\Vert^{r_n}_{r_n}\rightarrow\Vert f\Vert_p^p$, hence $\Vert f\Vert_{r_n}=\exp\left((1/r_n)\log\Vert f\Vert_{r_n}^{r_n}\right)\rightarrow\exp\left((1/p)\log\Vert f\Vert_p^p\right)=\Vert f\Vert_p$. This shows that $N_f$ is left-continuous at p. I don't know how to prove that $N_f$ is right-continuous

If we had $\Vert f\Vert_q<\infty$ for some $q>p$, the same kind of argument would have shown that $N_f$ is continuous at $p$.

Now, suppose $\Vert f\Vert_p=\infty$. Let $r_n\rightarrow p$. Since $|f|^{r_n}\rightarrow|f|^p$, then, by Fatou-Lebesgue, $\infty=\int|f|^p\leq \liminf\int|f|^{r_n}$, hence $\liminf\Vert f\Vert_{r_n}=\infty$, which shows that $\lim\Vert f\Vert_{r_n}=\infty=\Vert f\Vert_p$.

## marked as duplicate by Dan Rust, Nick Peterson, Thomas, Vedran Šego, M TurgeonOct 15 '13 at 0:39

What you are trying to prove is false, and your problem is exactly at the point where it fails.

Since the function $p \mapsto \|f\|_p$ is non-decreasing, either:

• $f \notin \mathbb{L}^\infty$,

or:

• $f \in \mathbb{L}^\infty$,

or:

• there exists some $p_0 \in [1, \infty]$ such that $f \in \mathbb{L}^p$ for all $p < p_0$ and $f \notin \mathbb{L}^p$ for $p > p_0$, minus the two cases above.

The first case is trivial. You already have solved the second case. For the third case, there are two sub-cases: either $f \in \mathbb{L}^{p_0}$ and $p_0 < + \infty$, or $f \notin \mathbb{L}^{p_0}$ and $p_0 > 1$. You have already solved the case $f \notin \mathbb{L}^{p_0}$, so the only reamining case is $f \in \mathbb{L}^{p_0}$ and $p_0 < + \infty$.

The problem is that, obviously, the function $p \mapsto \|f\|_p$ will be continuous on $[1, p_0]$ and on $(p_0, + \infty]$, but it won't be right-continuous in $p_0$: it jumps from a finite value to $+ \infty$.

The last question is: "Is this case possible"? It turns out it is, or in other words there are counter-examples to the proposition you are trying to prove. I'll let you find one of them.

Hint: to make things easier, take $p_0 = 1$ and $\Omega = [0, 1]$. Your goal is to find a non-negative function $f$ such that $f$ is integrable, but $f^p$ is not integrable for all $p > 1$.

• Wait, why is the function non-decreasing? – FTem Feb 1 '18 at 2:37
• @FTem: we work on a probability space, so using Jensen's or Holder's inequality, you can prove that $\|f\|_{\mathbb{L}^p} \leq \|f\|_{\mathbb{L}^p}$ whenever $p \leq q$. See e.g. en.wikipedia.org/wiki/Lp_space#Properties_of_Lp_spaces, section "Embeddings". – D. Thomine Feb 1 '18 at 12:12
• So is the function non-decreasing if $p\in(0,1)$? – FTem Feb 1 '18 at 12:21