Prove Intergal Question Without Calculating Integral

[daniel_i_l]

Homework Statement

let $$F(x) = {\frac{1}{x}}{\int^{x}_{0} e^{-t^2} dt}$$
Prove, without calculating the integral (but assuming that F exists in (0,infinity)) that:
a) the limit of F at 0 is 1.
b) for all x>0 $$\int^{x}_{0} e^{-t^2} dt <= x$$
x) supF((0,infinity)) = 1

Homework Equations

The Attempt at a Solution

a) I solved a using L'hopitals rule and the fundamental theorm.

b) First I defined a new function G where G(x) = F(x) for all x>0 and G(x) = 1 for x=0.
Now, I know that G is continues for all x>=0 and has a derivative for all x>0. So I have to prove that G'(x) < 0 for all x>0. I got that in order to prove that I have to prove that
$$\int^{x}_{0} e^{-t^2} dt > x e^{-x^2}$$ for all x>0. In order to do this I can define a new function $$H(x) \int^{x}_{0} e^{-t^2} dt - x e^{-x^2}$$. Now, H(0)=0 and for all x>0 H'(x) >0 and so H(x)>0 for all x>0 which means that G'(x) < 0 for all x>0 and so G(x) < 1 for all x>0 and since in this domain G(x)=F(x) we get that F(x)<1 for all x>0.
But this is a stronger result than what was asked for (=<), is my proof right?

c)We know from b that 1 is an upper bound of F in (0,infinity). Now, let's assume that there exists a number 0<c<1 so that c is also an upper bound. Since the limit of F at 0 is 1 we can find a D (delta) so that for all 0<x<D |F(x)-1| < (1-c)/2. This means that
F(x)-1 > -(1-c)/2 => F(x) > c which contradicts the fact that c is an upper bound!

Are those right? Especially (b) and (c).
Thanks.

 

[daniel_i_l]

Anyone? Please tell me if anything is unclear.
Thanks.

 

[Avodyne]

Sorry, the things you're supposed to prove all seem completely obvious to me, but I rarely know what constitutes a mathematically rigorous proof.

For (b), I would say e^(-t^2) <= 1 for all real t, so the its integral over any range will be <= the integral of 1 over that range. But I suppose I need to *prove* e^(-t^2) <= 1 for all real t first ...