Proving Uniform Convergence of f_n(x) in [a,b]

[Dick]

$$\mbox {Let } \epsilon<0,\ x_0 \in [a,b]$$

$$\lim_{n\rightarrow\infty} f_n(x)=f'(x)\ \Rightarrow\ \forall\ n>N\ |f_n(x_0)-f'(x_0)|< \frac {\epsilon}{3} \mbox { (*1)}$$

$$f_n(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta_n,\ |f_n(x)-f_n(x_0)|< \frac {\epsilon}{3} \mbox { (*2)}$$

$$f'(x) \mbox { is continuous in R } \Rightarrow\ \forall\ |x-x_0|< \delta,\ |f'(x_0)-f'(x)|< \frac {\epsilon}{3} \mbox { (*3)}$$

$$t_n=\min \{\delta_n, \delta \}$$ $$\Rightarrow\ \mbox {(*1) and (*2) and (*3) } \Rightarrow\ \forall\ |x-x_0|< t_n$$

$$|f_n(x)-f'(x)|=|f_n(x)-f_n(x_0)+f_n(x_0)-f'(x_0)+f'(x_0)-f'(x)| \leq |f_n(x)-f_n(x_0)|+|f_n(x_0)-f'(x_0)|+|f'(x_0)-f'(x)|< \frac {\epsilon}{3}+\frac {\epsilon}{3}+\frac {\epsilon}{3}<\epsilon$$

I have hard time to get sound intuition about this because the neighborhood around x is not fixed [changes for every n], and this makes me uncomfortable.




Now I'm trying to think about this. I'm not familiar with compactness theorem. [But I think Weierstrass Theorem can also help]

I have an idea now to play with $$t_n= \{x\ |\ \max_{[a,b]} |f_n(x)-f'(x)|\}$$, like I did with $$\mbox {x_0}$$ in the above proof.

You are absolutely right to be worried about the n dependence. Suppose you could show |f_n(x)-f_n(y)|<=M*|x-y| for some constant M, independent of n. Would that help?

 

[estro]

$$\mbox {Let } \epsilon<0,\ \ x_1,x_2 \in [a,b]$$

$$\lim_{n\rightarrow\infty} f_n(x)=f'(x)\ \ \ (*1)$$

$$(*1)\ \Rightarrow\ \forall\ n>N_1\ \ |f_n(x_1)-f'(x_1)|< \frac {\epsilon}{3}\ \ \ (*2)$$

$$(*1)\ \Rightarrow\ \forall\ n>N_2\ \ |f'(x_2)-f_n(x_2)|< \frac {\epsilon}{3}\ \ \ (*3)$$$$f'(x) \mbox{ is continuous in R } \Rightarrow\ \forall\ \ |x_1-x_2|<\delta\ \ |f'(x_1)-f'(x_2)|< \frac {\epsilon}{3}\ \ \ (*4)$$$$\mbox{(*2) and (*3) and (*4) } \Rightarrow\ \forall\ \ |x_1-x_2|<\delta\ and\ \forall\ n > N= \max\{ N_1,N_2 \}$$

$$|f_n(x_1)-f_n(x_1)|=|f_n(x_1)-f'(x_1)+f'(x_1)-f'(x_2)+f'(x_2)-f_n(x_2)| \leq |f_n(x_1)-f'(x_1)|+|f'(x_1)-f'(x_2)|+|f'(x_2)-f_n(x_2)|<\epsilon$$I'm almost sure this proof is right, but I only proved that f_n(x) is uniformly convergent on every open interval around some x in [a,b].

I'm thinking now to do what I did in the above proof with: x_1 = where f'(x) gets its maximum and x_2=where f'(x) gets its minimum.

 

[Dick]

Sorry, I'm not going to buy that. I don't think you've proved anything. What you have proved is that i) you really don't understand the issue and ii) that you are pretty good at ignoring good advice. |f_n(x)-f_n(y)|<=M*|x-y| for some constant M, independent of n. That is what you really need. Wouldn't that be helpful? Would you know how to prove it if you had to?