Home > Analysis, Real Analysis > Dini’s theorem and related problems

Dini’s theorem and related problems

Let { a<b} be two real numbers and { f_n :[a,b ]\rightarrow \Bbb{R}} a sequence of continuous functions which converge pointwise to a continuous function { f}.

1. Dini’s Theorem. Suppose that the sequence { (f_n)} has the property that { f_n \leq f_{n+1}}. Prove that the convergence { f_n \rightarrow f} is uniform.

2. Suppose that every function { f_n} is increasing. Prove that the convergence { f_n \rightarrow f} is uniform.

3. If every function { f_n} is convex on { [a,b]}, prove that the convergence { f_n \rightarrow f} is uniform on every compact interval in { (a,b)}.

Proofs: 1. Denote {g_n=f-f_n \geq 0} and {E_n = \{x \in [a,b] : g_n(x)<\varepsilon\}}. Notice that since {f_n} is increasing {g_n} must be decreasing, so {E_n \subset E_{n+1}}. Then each {E_n} is open and since {g_n} converges pointwise to zero it follows that {E_n} is a cover of {[0,1]}. Since {[0,1]} is compact, we can extract a finite subcover. The sets in this subcover are ordered by inclusion, so the greatest of them is the whole {[0,1]}. Therefore there exists {n_0} such that for {n \geq n_0} we have {E_n = [0,1]}. This implies the fact that the convergence is uniform.

2. The limit function {f} is obviously increasing. Pick {\varepsilon >0}. We can find {\alpha_x} such that {f(x+\alpha_x)-f(x-\alpha_x)\leq \varepsilon/3} (if we go outside the interval, extend {f,f_n} by making them constant outside {[a,b]}, keeping continuity. From the convergence of {f_n} to {f} we deduce the existence of {n_x} such that for every {n \geq n_x} we have

\displaystyle |f_n(x-\alpha_x)-f(x-\alpha_x)| \leq \varepsilon/3,|f_n(x+\alpha_x)-f(x+\alpha_x)| \leq \varepsilon/3.

Now if {y \in (x-\alpha_x,x+\alpha_x)} then {f_n(y) \geq f_n(x-\alpha_x) \geq f(x-\alpha_x)-\varepsilon/3} and {f_n(y) \leq f_n(x+\alpha_x) \leq f(x+\alpha_x)+\varepsilon/3}. Therefore {f_n(y)} can move in an interval of length {\varepsilon} containing {f(y)} which means {|f_n(y)-f(y)|\leq \varepsilon}.

Obviously the sets {(x-\alpha_x,x+\alpha_x)} cover {[0,1]}. Pick a finite subcover and {N_\varepsilon} the maximum of {N_x} for {x} in that subcover and we have that {|f_n(y)-f(y)|\leq \varepsilon } for every {n \geq N_\varepsilon}. Thus the convergence is uniform.

3. Let {[c,d]\subset (a,b)}. If {f_n} are convex then {f} is also convex. As a consequence, the functions

\displaystyle g_n(x)= \frac{f_n(x)-f_n(c-\varepsilon)}{x-(c-\varepsilon)},g(x)= \frac{f(x)-f(c-\varepsilon)}{x-(c-\varepsilon)}

are well defined and increasing for {\varepsilon>0} small enough. Furthermore the pointwise convergence {f_n \rightarrow f} implies the pointwise convergence {g_n \rightarrow g}. Apply the second part to conclude that the convergence {g_n \rightarrow g} is uniform. This implies that the convergence {f_n \rightarrow f} is also uniform.

  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: