2
$\begingroup$

Does this proof sound right? Thanks

Dini's Theorem: If $f$ and $f_n$ are continuous functions on $[a,b]$ such that $f_n \leq f_{n+1} \forall n \geq 1$ and $(f_n)$ converges to $f$ pointwise, then $(f_n)$ converges to $f$ uniformly.

My proof:

Let $g_n = f - f_n$. Fix $x_0 \in [a,b]$. Fix $\epsilon >0$ Use the pointwise convergence of $f_n$ to find an $N(x_0, \epsilon) \in \mathbb{N}$ such that $||f_k(x_0)-f(x_0)|| < \frac{\epsilon}{3} \forall k \geq N$. Since $f$ and $f_n$ are continuous, $\forall \epsilon > 0 \exists \delta(\epsilon, x_0) > 0$ such that $||f_n(x) - f_n(x_0)|| < \frac{\epsilon}{3}$ $\forall x \in [a,b]$ with $|x-x_0| < \delta$ and similarly for $f$. Then \begin{align} g_N(x) &= f(x) - f_N(x)\\ & \leq ||f(x) - f(x_0)|| + ||f(x_0)-f_N(x_0)|| + ||f_N(x_0) - f_N(x)|| \\ &\leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align}

Since $x_0$ was an arbitrary point, the function $g_n$ is uniformly convergent in every open interval around an arbitrary point $x_0 \in [a,b]$

Now suppose for contradiction that $f_n$ does not converge uniformly to $f$. $\lim_{n\to\infty}||f-f_n||_\infty = d > 0$ $\implies$ $\exists x_1 \in [a,b]$ such that $\lim_{n\to \infty}||f(x_1) - f_n(x_1)|| = d > 0$

Choose $\epsilon = \frac{d}{2}$ and choose a sequence $x_n \in [a,b]$ that converges to $x_1$.

$\lim_{n\to\infty}g_N(x_1)= f(x_1) - f_N(x_1) = d$ which is not less than epsilon and we have arrived at a contradiction.

This implies that $f_n \to f$ uniformly.

$\endgroup$
  • $\begingroup$ I don't think you need the contradiction part. $\endgroup$ – Paichu Dec 15 '18 at 2:15
  • $\begingroup$ Up and until the contradiction, I had my $N$ depend on $\epsilon$ and an arbitrary $x_0$. Is that enough for showing uniform convergence? $\endgroup$ – Amin Sammara Dec 15 '18 at 3:37
  • $\begingroup$ No. The requirement is that $N$ is independent of the choice of $x_0$. $\endgroup$ – xbh Dec 15 '18 at 3:45
0
$\begingroup$

Couple things:

  1. What's $\delta$? Please provide the exact meaning of symbols, otherwise it is confusing.
  2. You know that for each $x_0$ there exists $N(x_0, \varepsilon) \in \mathbb N$, then how could you deduce that $g_n \rightrightarrows g$ nearby each $x_*$?
  3. Where does the assumption $f_n \leqslant f_{n+1}$ applied?

Additional problem

  1. Is it necessary that such $x_1$ exists when $\lim \Vert f- f_n \Vert_\infty = d > 0 $?
$\endgroup$
  • $\begingroup$ 1) I edited I hope it’s better. $\endgroup$ – Amin Sammara Dec 15 '18 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.