Suppose $f: [0,1] \to \mathbb R$ is Riemann integrable on $[0,1]$, but not continuous on $[0,1]$. Let $$a_n = \left( \int_0^1 |f(x)|^n dx \right)^{\frac{1}{n}}$$ for $n \in \mathbb N$. Does $\lim_{n\to\infty} a_n$ exists? If it does, what is it equal to?

If $f$ is continuous, I know that $(a_n)$ converges to $M = \sup\{{|f(x)|: x \in [0,1]}\}$. But the proofs that I found for this case rely the continuity of $f$ to show that $\liminf_{n\to\infty} a_n \geq M$.

My idea is to use $M = \sup\{|f(x)|: x\in C\}$ instead, where $C\subseteq [0,1]$ is the set of all points at which $f$ is continuous. Then, if my reasoning is correct, there must be a $c \in C$ such that $\lim_{x\to c^+} f(x) = M$ or $\lim_{x\to c^-} f(x) = M$. Given an arbitrary $\varepsilon > 0$, perhaps I can then construct an interval $I \subset [0,1]$ such that $|f(x)| \geq M-\varepsilon$ for all $x \in I$. After that, the remaining parts should be similar to the continuous case. Nevertheless, I'm quite sure that there is some error in my line of reasoning, or maybe there is much more to be demanded for this argument to be complete.

From what I read, the term $a_n$ is actually $\|f\|_n$ (the $L^n$ norm), so $(a_n)$ should converge to $\|f\|_\infty$ as $n \to \infty$. However, my current understanding is limited to Riemann integration, without any knowledge whatsoever on measure theory and function spaces. Is there a way to prove the convergence of $(a_n)$ without resorting to measure theory, or even Lebesgue's Criterion?


2 Answers 2


A first observation is that you can use Hölder's inequality to show that the $a_n$ are increasing: $$ a_n^n = \int_0^1 |f(x)|^n \,dx$$ $$ \leq \left( \int_0^1 |f(x)|^{(n+1)} \,dx \right)^{n/(n+1)} \left( \int_0^1 1^{n+1 \,dx} \right)^{1/(n+1)} $$ $$ = \ (a_{n+1}^{n+1})^{n/(n+1)}=a_{n+1}^n, $$ so $a_n \leq a_{n+1}$.

Thus $\lim_{n \to \infty} a_n$ does always exist in a sense, but if $f$ is unbounded then the limit can (and will) be infinity. As an example you can look at something like $f(x) = x^{-1/10}$. Then $a_n$ is finite for $1 \leq n \leq 9$ but infinite for $n \geq 10$.

  • $\begingroup$ Thank you for your answer. Correct me if I'm wrong, but isn't it the case that every Riemann integrable function on a closed interval is bounded? If that's the case, then $(a_n)$ would always converge to a finite real number. Or am I missing some underlying assumptions here? $\endgroup$ Oct 7, 2020 at 9:54
  • 1
    $\begingroup$ @cosmic_philosopher that depends on whether you are considering the proper Riemann integral, in which case you are right, or the possibly improper riemann integral $\endgroup$
    – Caffeine
    Oct 7, 2020 at 12:20

Yes. In fact, even for Lebesgue integrable functions $f$ we have $$ \lim_{n \to \infty} \left( \int_0^1 |f(x)|^n dx \right)^{\frac{1}{n}} = \text{ess.sup} |f| $$ where the "essential supremum" $\text{ess.sup} |f|$ is the least $A$ such that $\{x:|f(x)| > A\}$ has Lebesgue measure zero. Sometimes $\text{ess.sup} |f| = +\infty$. If $f$ is Riemann integrable, then $f$ is bounded, so $\text{ess.sup} |f|$ is finite. Also note, if $f$ is continuous, then $\text{ess.sup} |f| = \max \{|f(x)| : x \in [0,1]\}$.

  • $\begingroup$ Thanks, fixed.. $\endgroup$
    – GEdgar
    Oct 7, 2020 at 12:29

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