Differentiability of a Twice Differentiable Function

[OXF]

Homework Statement

Let g:R->R be a twice differentiable function satisfying g(0)=g'(0)=1 and g''(x)=g(x)=0 for all x in R.
(i) Prove g has derivatives of all orders.
(ii)Let x>0. Show that there exists a constant M>0 such that |g^n(Ax)|<=M for all n in N and A in (0,1).

Homework Equations

Possibly Rolle's Theorem, Cauchy's Mean Value theorem.

The Attempt at a Solution

 

[hunt_mat]

I think you have written the definition down wrong, as if g''(x)=g(x)=0 for all x in the real numbers, then take x=0 and you have a contradiction in your definition.

 

[OXF]

Ah, I meant g''(x)+g(x)=0, apologies.

 

[hunt_mat]

you know that g(x) is differentiable, so we know g''(x)=-g(x), so we know that as g is differentiable then that shows that:

<br /> \lim_{h\rightarrow 0}\frac{g&#039;&#039;(x+h)-g&#039;&#039;(x)}{h}=-\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}<br />

So as h tends to 0, the limit on the RHS exists for all x and therefore limit on the LHS must exist for all x too. Carrying this on shows that g(x) is infinitely differentiable.