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

## Dini’s theorem and related problems

Let ${ a 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.