Proving That f is Analytic with All Orders of Derivatives

[futurebird]

f is analytic and has all orders of derivatives.

I'm trying to prove something here but doing this makes me uneasy... is there anything wrong with saying:

If $$|f(z)| \leq |z^{n}|$$ then,

$$|f'(z)| \leq |nz^{n-1}|$$ so

$$|f''(z)| \leq |n(n-1)z^{n-2}|$$
.
.
.

$$|f^{n}(z)| \leq n!z^{0}$$

$$|f^{n}(z)| \leq n!$$

$$|f^{n+1}(z)| \leq 0$$

$$|f^{n+1}(z)| = 0$$

I can find no support for taking the derivative of the inequality like this in my calc book.

 

[eok20]

Taking the derivative of both sides of the inequality is not justified. For example, consider f(x) = sin(e^x) for x > 1. |f(x)| < x^2 but it is not that case that |f'(x)| = |e^x cos(e^x)| < 2x (for example, when x=ln(2*pi)). However, the fact that youre comparing an analytic function to a polynomial in z makes me think that Taylor's theorem may help you.

 

[futurebird]

Thanks. IT just seemed way too easy.

How do they solve differential inequalities then, I wonder.