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 $f(x) = 0 \ \forall x \in [a, b]$

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    $\begingroup$ $f_n(x) = x^n$ 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

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 $[a,b]$.

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  • $\begingroup$ @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

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