0
$\begingroup$

Let's consider the metric space $C^0([a, b]), \ a, b, \in \Bbb R$ (the set of continuous functions defined on the interval $[a, b]$) equipped with the norm $\lvert\lvert f \rvert\rvert = \int_a^b \lvert f \rvert (x) dx$.

Let $(f_n)_{n\geq1}\subset C^0([a, b])$ be a convergent sequence in $(C^0([a, b]), \lvert\lvert \cdot \rvert\rvert )$ , i.e. $\exists f_{*} \in C^0([a, b])$ such that $\lim \limits_{n\to\infty} \lvert\lvert f_n - f_{*}\rvert\rvert = \lim \limits_{n\to\infty} \int_a^b \lvert f_n - f_{*} \rvert (x) dx = 0.$

Does that imply $\lim \limits_{n\to\infty} \lvert\lvert f_n - f_{*}\rvert\rvert_{\infty} = 0$ ?

I thought about that since if $f \in C^0([a, b])$ and $\int_a^b \lvert f \rvert (x) dx = 0$, then $f(x) = 0 \ \forall x \in [a, b]$

$\endgroup$
  • 3
    $\begingroup$ $f_n(x) = x^n$ on $[0,1]$? $\endgroup$ – user251257 Oct 25 '16 at 17:35
  • $\begingroup$ No, it doesn't imply that. Think about $ x^n $ as mentioned above. It just means that as $ n $ goes to infinity the x-spread of points which aren't uniformly convergent goes to zero. $\endgroup$ – QuantumFool Oct 25 '16 at 17:36
  • $\begingroup$ @QuantumFool So this behaviour could happen in the interior of $[a, b]$ with a fitting function sequence ? Can we deduce anything interesting from the convergence with the integral norm ? $\endgroup$ – Desura Oct 25 '16 at 17:57
2
$\begingroup$

Convergence in $L^1$ does not imply convergence in $L^\infty$. If we take an enumeration $q_1,q_2,q_3,\ldots$ of the rational numbers in $[a,b]$ and consider $$ f_n(x) = \exp\left(-n|x-q_n|\right)\in C^0([a,b]) $$ then in $L^1(a,b)$ we have $f_n(x)\to 0$, but for any $n\geq 1$ we have $f_n(q_n)=1$, hence we cannot have uniform convergence. On the other hand, by the relations between modes of convergence we have $L^1 -\! -\!\rightarrow\text{AU}$, hence there is a subsequence $\{f_{n_k}\}_{k\geq 0}$ that is convergent to $0$ almost uniformly on $[a,b]$.

$\endgroup$
  • $\begingroup$ @DominiqueR.F.: let we consider a sequence $q_{n_k}\to\eta\in[a,b]$. Then for every neighbourhood $U$ of $\eta$ with radius $\varepsilon$, $f_{n_k}$ uniformly converges to $0$ on $[a,b]\setminus U$, i.e. $f_{n_k}$ is almost everywhere uniforly convergent to zero on $[a,b]$. $\endgroup$ – Jack D'Aurizio Oct 25 '16 at 20:49
  • $\begingroup$ @DominiqueR.F.: oh, I see, you're right. I wrote "almost everywhere" but I should have written "almost uniformly". Now fixed. Anyway, $L^1 -\!-\!\rightarrow\text{AE}$ also holds. $\endgroup$ – Jack D'Aurizio Oct 25 '16 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.