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I'd like to prove the following:

If $f_j$ are continuous functions on a compact set $K$, and $f_{1}(x) \leq f_{2}(x) \leq \dots$ for all $x \in K$, and the $f_j$ converge pointwise to a continuous function $f$ on $K$ then in fact the $f_j$ converge uniformly to $f$ on $K$.


Let $g_{j}(x) = f(x) - f_{j}(x)$ for all $j$. Then, since $f_j \rightarrow f$ pointwise, we see $g_j \rightarrow 0$ pointwise.

Now, let $\varepsilon > 0$ . And examine { $x \in K : g_{j}(x) < \varepsilon$ }.

I've been told that the next step should be to show that { $x \in K : g_{j}(x) < \varepsilon$ } is equal to the intersection of $K$ with some open set $U_j$. But I'm not certain why this is true?

Advice? Insight?

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If $h$ is a continuous function, what can you say about $h^{-1}(-\infty,\varepsilon)$? – Davide Giraudo Feb 28 '12 at 22:14

You are almost there. What you need to do is to consider the sets $$ H_j=\{x:\ g_j(x)<\varepsilon\},\ \ \ j\in\mathbb{N}. $$ These sets are increasing, i.e. $H_{j}\subset H_{j+1}$. The continuity of $f$ guarantees that $g_j$ is continuous for all $j$, and so $H_j$ is open, since $H_j=g_j^{-1}(-\infty,\varepsilon)$.

For any $x\in K$, $g_j(x)\to0$, so there exists $j$ such that $x\in H_j$. This means that $$ K\subset\bigcup_j H_j, $$ so the $H_j$ make an open cover of $K$. By compactness, $K$ is contained in the union of a finite family $H_{j_1},\ldots,H_{j_r}$, $j_1<\ldots< j_r$. But then $$ K\subset H_{j_1}\cup\cdots\cup H_{j_r} = H_{j_r}. $$ In other words, for any $x\in K$, for any $j\geq j_r$, $g_j(x)<\varepsilon$; so the convergence is uniform.

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Thanks, you helped with my homework too. :) – Ory Band Jun 15 '12 at 14:43

Adding a slighly different proof (same idea though), mainly for the sake of documenting it for myself and so I won't forget the theorem:

First let $t \in K$ be fixed. Note that monotonicity of the sequence $f_j(t)$ and its convergence to $f(t)$ implies that $f_j(t) \leq f(t)$.

For $\delta >0$, we define $U_\delta(t) = \left\{\tilde{t} \in K: |\tilde{t}-t|< \delta\right\}$.

Now pick $\varepsilon > 0$, and pick $N_t$ and $\delta_t > 0$ such that:

  • $$\left|f_n(t) - f(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; n\geq N_t$$
  • $$\left|f_{N_t}(\tilde{t}) - f_{N_t}(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; \tilde{t} \in U_{\delta_t}(t)$$
  • $$\left|f(\tilde{t}) - f(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; \tilde{t} \in U_{\delta_t}(t)$$

These exist by the pointwise convergence and the continuity of $f_{N_t}$ and $f$.

It now holds for every $n \geq N_t$:

$$ \sup_{\tilde{t} \in U_{\delta_t}(t)} | f(\tilde{t}) - f_n(\tilde{t})|=\sup_{\tilde{t} \in U_{\delta_t}(t)} f(\tilde{t})-f_n(\tilde{t}) \leq \sup_{\tilde{t} \in U_{\delta_t}(t)} f(\tilde{t}) - f_{N_t}(\tilde{t}) \leq f(t) - f_{N_t}(t) + 2\frac{\varepsilon}{3} \leq \varepsilon $$

Finally, notice that $\displaystyle\bigcup_{t \in K} U_{\delta_t}(t) \supset K$ is an open cover of $K$. Compactness yields $t_1, \dotsc, t_r \in K$ such that $$\displaystyle\bigcup_{i \in \{1,\dotsc,r\}} U_{\delta_{t_i}}(t_i) \supset K$$

Now just take $N:= \max\{N_{t_1},\dotsc, N_{t_r}\}$ to finish the proof.

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