I am in need of results transforming pointwise convergence of functions into uniform convergence. Since I wasn't satisfied with Dini's theorems, I had to prove the following result:

Let $K$ be a topological compact space, and $C(K)$ the space of continuous functions from $K$ to $\mathbb{R}$, which is a Banach space once endowed with the uniform norm $\Vert \cdot \Vert_{C(K)}$.

Let $(f_i)_{i\in I}$ be a net in $C(K)$, and $f \in C(K)$. We make the two following assumptions:

1) For any net $(x_i)_{i \in I}$ converging to some $x$ in $X$, we have $f_i(x_i) \rightarrow f(x)$ $\ $ (in particular, $f_i$ is pointwise convergent to $f$),

2) The net $(f_i)_{i \in I}$ is bounded in $C(K)$.

Then $f_i$ converges to $f$ uniformly on $K$.

So my question is in the title : have you ever seen this result? If so, do you have a reference where I could find it? It does not seem to be close to Dini's theorems, since no monotony is involved here. Clearly the main ingredients here are the compactness of $K$ and hypothesis 1). The latter is stronger that pointwise convergence, and recalls somehow the $\Gamma$-convergence.

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Just in case, I put here the proof of the result :

If we write $r_i:=\Vert f_i - f \Vert_{C(K)}$, we just need to show that $r_i$ tends to zero. Because of hypothesis 2), we know that $(r_i)_{i \in I}$ is bounded in $\mathbb{R}$, hence relatively compact. So, all we need to prove is that any limit point of $(r_i)_{i \in I}$ is zero.

Let $r \in \mathbb{R}$ be such a limit point, then there exists a subnet (that I note abusively:) $(r_j)_{j \in I}$ such that $r_j$ tends to $r$. From the definition of $\Vert \cdot \Vert_{C(K)}$, we have $r_j=\sup\limits_{x \in K} \vert f_j(x) - f(x) \vert$ for all $j \in I$. By hypothesis, $f_j - f$ is continuous on the compact $K$, so by the extreme value theorem, there exists some $x_j \in X$ such that $r_j = \vert f_j(x_j) - f(x_j) \vert$.

The net $(x_j)_{j \in I}$ lies in the compact $X$, so there exists a subnet $(x_k)_{k \in I}$ converging in $X$ to some $x$. In particular, because of hypothesis 1), we obtain $r_k=\vert f_k(x_k) - f(x_k) \vert \rightarrow 0$. So we can conclude that $r=0$.


You don't need to assume hypothesis (2). If $f_i$ does not converge uniformly to $f$, there is $\epsilon > 0$ and a subnet $J$ such that $\|f_j - f\| \ge \epsilon$ for $j \in J$, therefore $x_j$ such that $|f_j(x_j) - f(x_j)| \ge \epsilon$. Take a subnet $\{x_k\}_{k \in K}$ of $\{x_j\}_{j \in J}$ that converges to some $x$ and you find $|f(x) - f(x_k)| \ge \epsilon$, contradicting continuity of $f$.

  • $\begingroup$ Nice remark, thank you ! $\endgroup$ – Guillaume Garrigos Jun 12 '15 at 5:40

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