Let's first state the Ascoli-Arzela theorem in $C([a,b])$.

Theorem (Ascoli-Arzela) Let $(f_n)_{n\in\mathbb{N}}\in C([a,b])^\mathbb{N}$ such that:

  • (uniform boundedness) there exists $M>0$ such that $\sup\limits_{n\in\mathbb{N}}\Vert f_n\Vert_{\infty}\leqslant M$
  • (uniform equicontinuity) for all $\varepsilon>0$, there exists $\delta_\varepsilon>0$ such that for all $(x,y)\in[a,b]^2$, $$ \vert x-y\vert\leqslant\delta_\varepsilon\Longrightarrow \sup_{n\in\mathbb{N}}\vert f_n(x)-f_n(y)\vert\leqslant\varepsilon $$ Then there exists a sub-sequence $(f_{\sigma(n)})_{n\in\mathbb{N}}$ and a function $f\in C([a,b])$ such that $(f_{\sigma(n)})_{n\in\mathbb{N}}$ converges uniformly to $f$, that is $$ \forall\varepsilon>0,~~ \exists N_\varepsilon\in\mathbb{N}, ~~\forall n\geqslant N_\varepsilon, ~~\Vert f-f_n\Vert_{\infty}<\varepsilon $$

The problem is about making the assumptions weaker and still have some results. Precisely:

The problem Let $(f_n)_{n\in\mathbb{N}}\in C([a,b])^\mathbb{N}$ such that:

  • (pointwise boundedness) for all $x\in[a,b]$, there exists $M_x>0$ such that for all $n\in\mathbb{N}$, we have $\vert f_n(x)\vert\leqslant M_x$
  • (simple equicontinuity) for all $x\in[a,b]$ and $\varepsilon>0$, there exists $\delta_{x,\varepsilon}$, such that for all $y\in[a,b]$, $$ \vert x-y\vert\leqslant\delta_{x,\varepsilon}\Longrightarrow \sup\limits_{n\in\mathbb{N}}\vert f_n(x)-f_n(y)\vert\leqslant \varepsilon $$ Show that $(f_n)_{n\in\mathbb{N}}$ is uniformly bounded , that is there exists $M>0$ such that for all $n\in\mathbb{N}$, we have $$ \Vert f_n\Vert_\infty\leqslant M $$

My progress If we pick $x\in [a,b]$ and $n\in\mathbb{N}$, we have for all $y\in [a,b]$, $$ \vert f_n(x)\vert\leqslant\vert f_n(x)-f_n(y)\vert+\vert f_n(y)\vert $$ We have $\vert f_n(y)\vert\leqslant M_y$ (the bound being independent of $n$), and for $y$ close enough to $x$, the second assumption shows that $\vert f_n(x)-f_n(y)\vert\leqslant C$ (the bound being also independent of $n$). Thus, for all $x\in [a,b]$, for all $n\in\mathbb{N}$, we can bound $\vert f_n(x)\vert$ by a constant independent of $n$ but dependent on an element in the neighborhood of $x$.

But I need to show that this bound does not depend on $x$ to finish the proof (I think). What is the way to show this ? An intuition would be that the boundedness "spreads" across all the closed interval, but it seems the neighborhood can shrink at each step we are trying to "spread" the bound.

Any hint would be appreciated, thanks !

PS: is a sequence that has simple equicontinuity in $C([a,b])$ also has uniform equicontinuity ?


The intervals $(x-\delta_{x,\epsilon},x+\delta_{x,\epsilon})$ form an open cover of $[a,b]$. By compactness there is a finite subcover $(x_i-\delta_i, x_i+\delta_i), 1 \leq i \leq N$. For $x$ there exists $i$ such that $x \in (x_i-\delta_i, x_i+\delta_i)$ and $|f_n(x)| \leq \epsilon +|f_n(x_i)|$. By hypothesis $|f_n(x_i)|$ is bounded for each $i$. Can you finish?

The answer to your last question is YES. It is again a compactness argument.

  • $\begingroup$ Thank you for your answer, I am able to finish the proof ! $\endgroup$ – Flewer47 May 9 '20 at 10:20

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