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My question has two parts. Both are related to a family of functions $(f_r)_{r\in[0, +\infty[}$.

1. Properties of uniformly convergent families.

Assuming $f_r \rightrightarrows f$ as $r\to +\infty$, do any of the theorems from sequences still apply? For instance:

  • Does the continuity of every $f_r$ imply that of $f$?
  • If every $f_r$ is differentiable, and the derivatives converge uniformly, do we get locally uniform convergence to $f$ and derivative/limit permutation?

2. Can the Arzelà–Ascoli theorem be used to prove the convergence of a family?

I am inclined to say yes. Below is my attempt at a proof. Let me know if I made some mistake, or if the claim is outright false.

Theorem.

If $(f_r)$ is a family of uniformly bounded, uniformly equicontinuous functions on $\mathbb{R}^n$, such that $f_r \to f$ pointwise for some continuous function $f$, then $f_r\to f$ locally uniformly.

proof. Let $K$ be a compact set and $\varepsilon > 0$, by equicontinuity, there is a $\delta > 0$ such that: $$ \forall r\ge 0, \ \|x-y\|<\delta \implies |f_r(x) - f_r(y)| < \varepsilon/3. $$

Now let $$ K \subset \bigcup_{i=1}^N B_\delta(x_i), $$ be a finite cover of $K$ with balls of radius $\delta$ (exists by total boundedness). By pointwise convergence, we find an $R > 0$ such that: $$\forall 1 \le i \le N,\ r>R \implies |f_r(x_i) - f(x_i)| < \varepsilon/3.$$ Let $x \in K$. By construction $x\in B_\delta (x_i)$ for some $x_i$. The triangle inequality yields: $$ |f_r(x) - f(x)| \le |f_r(x) - f_r(x_i)| + |f_r(x_i) - f(x_i)| + |f(x_i) - f(x)|. $$

Now we have: $|f_r(x) - f_r(x_i)| < \varepsilon/3$ by equicontinuity, $|f_r(x_i) - f(x_i)| < \varepsilon/3$, by pointwise convergence, and we can choose $\delta$ such that $|f(x_i) - f(x)| \lt \varepsilon / 3$ by continuity of $f$.

It follows that $$ |f_r(x) - f(x)| < \varepsilon \implies \forall r > R, \ \sup_{K}{|f_r - f|} > \varepsilon, $$ from which the result follows. $\square$

Any help with either part of the question is greatly appreciated.

0sharp
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1 Answers1

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  1. You can simply use the results for sequences: if $f_r$ converges uniformly to $f$, then the sequence $(f_n)_{n \in \mathbb{N}}$ also converges uniformly to $f$, and thus you may apply the corresponding results for sequences.

  2. You can also argue by contradiction, again using the results for sequences. Suppose the convergence is not uniform on some compact set $K$. Then there exists $\epsilon > 0$ such that for every $n$ there are $r_n \to \infty$ and $x_n \in K$ with $$ |f_{r_n}(x_n) - f(x_n)| > \epsilon. $$ However, by the Arzelà–Ascoli theorem for sequences, there exists a subsequence of $(f_{r_n})$ that converges uniformly to $f$ on $K$, leading to a contradiction.

Indeed, the Arzelà–Ascoli theorem in one of its most general forms is precisely a characterization of equicontinuous and (locally) uniformly bounded families of functions.

Daniel Smania
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