# Ngô Quốc Anh

## March 27, 2008

### Liên tục đều vs. Tính khả vi…

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 20:01

Let $f$ be continuous and bounded on $\mathbb{R}$ function such that $\sup\limits_{x\in\mathbb{R}}|f(x + h) - 2f(x) + f(x - h)|\to0,\quad h\to0.$

Does it follows that $f$ is uniformly continuous on $\mathbb{R}$?

Solution. Suppose for example that $|f|$ is bounded and that $f$ is not uniformly continuous. Then one can find $\epsilon>0$ such that there exists a sequence $(x_n,y_n)$ satisfying $|x_n-y_n| \to 0$ and $|f(x_n)-f(y_n)| \geq \epsilon$. Now, take $h_n$ such that $\sup_{x \in \mathbb{R}, |h| \leq h_n } \{f(x-h)-2 f(x)+f(x+h)\} \leq \frac{\epsilon}{n}$

So if $x_n satisfies $\delta_n = y_n-x_n \leq h_n$ and $f(y_n) \geq f(x_n) + \epsilon$, then it is easy to see that: $f(x_n+ (k+1) \delta_n)-f(x_n + k \delta_n) \geq \epsilon (1 - \frac{k}{n})$

for $0 \leq k \leq n-1$. Hence $f(x_n+ n \delta_n) - f(x_n) \geq \epsilon(\frac{n}{n} + \frac{n-1}{n}+ \ldots + \frac{1}{n})=\frac{(n+1) \epsilon}{2}$,

which shows that $f$ cannot be bounded, a contradiction.

More. Suppose $f$ continous function on $]a,b[$

Suppose there exist $C>0$ such that $|\frac{f(x+h)+f(x-h)-2f(x)}{h}|\leq C$

each time it is possible (in french “chaque fois que cela a un sens”)

1/ prove $f$ is bounded on $]a,b[$

2/ Suppose $x,x+h \in ]a,b[$ with $h>0$

Prove for any $n \in N$ $|f(x+\frac{h}{2^n})-f(x)|\leq \frac{C.n.h}{2^{n+1}} + \frac{|f(x+h)-f(x)|}{2^n}$

Prove there exist $C'>0$ such that for any $d>0$ small $\sup_{|x-y|\leq d}|f(x)-f(y)| \leq C'.d.\ln(1/d)$

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