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THEOREM: Suppose $\{f_n\}$ is a sequence of continuous functions from $[a,b]$ to $\Bbb R$ that converge pointwise to a continuous function $f$ over $[a,b]$. If $f_{n+1}\leq f_n$, then convergence is uniform.

Then, why is the continuity of the the functions $f_i$'s important for the theorem?

Almost all textbooks seems to use property of compactness which I will get to learn only in Metric Spaces chapter. Can anyone please give me an alternate reasoning?

Attempt: If each $f_k$ is continuous in $[a,b]$, it means it is bounded as well. Let the infimum of $f_k$ b3 denoted as $m_k$ and supremum as $M_k$ in $[a,b]$. Then, since the sequence of functions is given to be monotonically increasing, this means:

$m_k < m_{k+1} < .......< m_n <...$ and $M_k < M_{k+1} < .......< M_n < ...$

How do I proceed next?

Edit : Unless $f_n(x)$ is continuous at every point, we might not be able to infer the very definition of uniform convergence which states that a uniformily convergent sequence of functions $f_i$ such that $lim ~f_i = f$, then unless each $f_i$ is continuous, we won't be able to find the value of $f_i(x)$ at every point $x$ and hence won't be able to write the following definition of uniform convergence "

$\forall \epsilon >0, \exists ~m \in N$ s.t $f(x)-f_i(x) < \epsilon ~~\forall~~ n\geq m ~~\forall x ~\in ~ I$ where $I$ is the given interval

Would this be correct??

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    $\begingroup$ $ f_n$ is decreasing so use $f_n - f$ Also at this point m depends on x because you are still trying to prove uniform convergence. $\endgroup$ – RRL May 14 '14 at 22:33
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Compactness cannot be avoided.

Let $\varepsilon>0$ and $x \in [a,b]$.

Since $f_n(x)$ converges pointwise to $f(x)$, there exists $N_x$ such that $$|f_n(x)-f(x)|<\varepsilon/3 \quad \text{for} \quad n \ge N_x$$ Also, since $f$ and $f_{N_x}$ are continuous, there is an open neighborhood $U_x$ of $x$ such that $$|f(x)-f(y)|<\varepsilon/3\quad \text {and} \quad |f_{N_x}(x)-f_{N_x}(y)|<\varepsilon/3 \quad \text {for} \quad y \in U_x \cap[a,b]$$ Moreover, by monotonicity, one has $$f(x) \le f_n(x) \le f_m(x) \quad \text {for} \quad n\ge m$$ From compacteness of $[a,b]$ it follows that there are a finite number of points $x_i \in [a,b]$ such that the $U_{x_i}$ cover $[a,b]$.
Let $\bar N$ be the maximum of the $N_{x_i}$.
Then, for any $x \in [a,b]$, $\,x$ belongs to some $U_{x_i}$ and $$|f_n(x)-f(x)|\le$$$$|f_{N_{x_i}}(x)-f(x)| \le$$$$|f_{N_{x_i}}(x)-f_{N_{x_i}}(x_i)|+|f_{N_{x_i}}(x_i)-f(x_i)|+|f(x_i)-f(x)|<\varepsilon$$ for $n \ge \bar N$ .

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  • $\begingroup$ Thank you for the answer :) $\endgroup$ – MathMan May 18 '14 at 18:52
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As far as I know, there isn't a way to prove this theorem without using properties of compactness. In fact, compactness of domain is necessary for the statement to be true.

So, let's just develop those relevant properties here, in $\mathbb{R}$ instead of in metric spaces. Hopefully working through these definitions and exercises will help.

Definition: A limit point of a set $S \subset \mathbb{R}$ is a point $p$ such that, for every $\epsilon>0$, there is $x\in S$ with $|p-x|<\epsilon$.

Definition: A subset $S\subset \mathbb{R}$ is open if, for all $x\in S$, there is some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subset S$.

Definition: A subset $S\subset \mathbb{R}$ is closed if it contains all of its limit points.

Exercise 1: Prove that a set $A\subset \mathbb{R}$ is open if and only if $A^c$, its complement, is closed.

Exercise 2:

a) Let $\{A_n\}$ be a sequence of open sets. Prove $\bigcup_{n=1}^{\infty} A_n$ is itself open.

b) Let $\{A_n\}$ be a sequence of closed sets. Prove $\bigcap_{n=1}^{\infty} A_n$ is itself closed.

Exercise 3: Prove that $f:\mathbb{R} \to \mathbb{R}$ is continuous if and only if $f^{-1} (U)$ is open for every open subset $U \subset\mathbb{R}$. What does this tell you about $f^{-1} (V)$ for closed sets $V\subset\mathbb{R}$?

Definition: An open covering of a set $S\subset \mathbb{R}$ is a collection of open sets $\{U_\alpha\}$ (countable or uncountable), such that $S\subset \bigcup U_\alpha$. A finite subcovering of $\{U_\alpha\}$ is some subcollection $\{U_1, \cdots, U_n\}$ of $\{U_\alpha\}$ such that $S \subset \bigcup_{i=1}^{n} U_i$.

Definition: A set $S \subset\mathbb{R}$ is compact if every open covering of $S$ contains a finite subcovering of $S$.

Exercise 4: Let $S\subset \mathbb{R}$ be compact. Prove that every infinite subset of $S$ has a limit point.

Theorem (Heine-Borel): A set $S\subset\mathbb{R}$ is compact if and only if it is closed and bounded. This is hard to prove, but think about why it should be true.

Proof of Dini's Theorem:

Fix $\epsilon> 0$. Let $g_n = f_n - f$ for all $n$. Since $\{f_n\}$ is decreasing, we see $g_{n+1} \le g_n$ for all $n$, and also $g_n \ge 0$ for all $x\in [a,b]$. In addition $g_n (x) \to 0$ as $n\to\infty$ for all $x\in [a,b]$ (by pointwise convergence of the $\{f_n\}$).

Now, consider the sets ${g_n}^{-1} ([\epsilon, \infty))$. First, check that $[\epsilon, \infty)$ is a closed set. By Exercise 3, we then have that ${g_n}^{-1} ([\epsilon, \infty))$ is also closed, for all $n$. Thus, ${g_n}^{-1} ([\epsilon, \infty))$ is compact for every $n$ (why?). Finally, note that $${g_{n+1}}^{-1}([\epsilon, \infty)) \subset {g_n}^{-1}([\epsilon,\infty))$$

The statement that the $\{f_n\}$ converge uniformly to $f$ is equivalent to the statement that ${g_n}^{-1} ([\epsilon, \infty)) = \emptyset$ for some $n$ (why?). So suppose this is not true. Then for each $n$, we can pick $x_n \in {g_n}^{-1} ([\epsilon, \infty))$. By Exercise 4, the sequence $\{x_i\}$ has a convergent subsequence $\{x_{i_k}\}$. Say that $x_{i_k} \to x$ as $k\to\infty$. We must have $x\in {g_n}^{-1} ([\epsilon, \infty))$ for all $n$ (why? Use Exercise 2). Thus, $g_n (x) \ge \epsilon$ for all $n$. This contradicts the fact that $g_n (x) \to 0$ pointwise.

Conclusion: The $\{f_n\}$ converge uniformly to $f$.

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    $\begingroup$ Thank you for your answer :) $\endgroup$ – MathMan May 15 '14 at 4:16

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