Analysis Question-differentiabillity, continuity

[JohnnyBG00d]

Analysis Question--differentiabillity, continuity

Homework Statement

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a $C^\infty$ function which satisfies the equation $f''(x)=-x^2f(x)$ along with $f(0)=1$, $f'(0)=0$. Prove that there is an $a>0$ such that $f(a)=0$. Do not use any results from differential equations. Thank you.

Homework Equations

The Attempt at a Solution

Since f is continuously differentiable there is a $(0,\delta)$ interval in which f is concave down. If we can show f is also decreasing then it follows that f must cross the x-axis before changing concavity and increasing because there can be no cusps as f is differentiable everywhere. I have no idea if that is the right track. Thank you.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[micromass]

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[klondike]

If we can show f is also decreasing then it follows that f must cross the x-axis before changing concavity and increasing

Not true. A counterexample sin(x)+2 goes between convex and concave without crossing x.

I have no idea if that is the right track. Thank you.

Perhaps it'd be easier to prove by contradiction.
f(0)=1>0, by continuity, in the immediate neighborhood xE[0,c], f(x)>0. Let's (erroneously) assume f(x)>0 for all x>0. since f''(x)<0, f(x) concave down, it follows that
f(x)<f(c)+f'(c)(x-c) for all x>0 and c>0
You can prove f'(x)<0 for all x>0 (if the assumption is true), hence f'(x)=-|f'(x)|.

Now, let x=ζ=c+f(c)/|f'(c)|, it leads to
f(ζ)<0 where ζ>c>0. Now this contradicts with the assumption that f(x)>0 for all x>0.

Not pretty, just to bounce some idea.