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<y_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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