suppose f:R->R is an infinitely differentiable function with |f(0)|>M
and |f'| interval (0,a) |f|>=M-Ta.

Now suppose f:R->C is an infinitely differentiable function with |f(0)|

(0,a) where |f| is bounded away from zero?

On 16-05-2008 7:23, novis wrote:

Sure. Use the mean value theorem for complex-valued functions. It states
that if f:[a,b] ---> C is differentiable, then

|f(b) - f(a)| <= sup{ |f'(c)| | a <= c <= b }.|b - a|.

Best regards,

Jose Carlos Santos

if |f'|>N>0 can similarly be said from mean value theorem that
|f(b)-f(a)|>N|b-a| ????
Thanks

In article
56g2000hsm.googlegroups.com>,
novis gmail.com> wrote:

For C, no. Example f(t) = exp(it), a=0, b=2 Pi.
Then f(b)=f(a), but |f'(t)|=1 for all t.

if (a,b) in C^2, let g(t)=|f(a+t(b-a))|for t in R

f(0)<>0 and f is continuous hence g(t)<>0 for t "near" 0 and so is g
differrentiable and you can use the mean value theorem.

You will find that ||f(x)|-|f(0)||<=T|x-0|.

But ||f(x)|-|f(0)||>=|f(0)|-|f(x)| hence |f(x)|>=M-T|x| on an interval
containing 0.