# Using proof by contradiction to prove Dini's theorem

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.

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

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?
1. Is it necessary that such $$x_1$$ exists when $$\lim \Vert f- f_n \Vert_\infty = d > 0$$?