# Let $f:[-1,1] \to \mathbb{R}$ be differentiable 3 times, prove $\exists M>0 \ , \ s.t \ f(x) \le Mx^2$

Let $$f:[-1,1] \to \mathbb{R}$$ be differentiable 3 times, let $$f(0)=0$$ and $$f(x) \ge 0 \ \forall x \in [-1,1]$$.

Prove: $$\exists M>0 \ , \ s.t \ f(x) \le Mx^2$$.

I separated the proofs to two cases:

1. $$f(x)=0 \forall x \in [-1,1]$$, so we are done.
2. $$\exists x \in [-1,1] \ s.t \ f(x) \ne 0$$, so we can say that $$x=0$$ is an extremum, so $$f'(0)=0$$.

I have expanded the function using Taylor expansion: $$f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2}+\frac{f'''(c_x)x^3}{6}=\frac{f''(0)x^2}{2}+\frac{f'''(c_x)x^3}{6}$$ where $$c_x \in (-1,1)$$.

I don't know how to continue...

You're on the right track, but I believe all you need to use is that $f$ is $C^2$ ($f''$ is continuous). Use the first degree Taylor polynomial with remainder to get $$f(x)=f(0)+f'(0)x+\frac12 f''(c_x)x^2 \quad\text{for some } x_c \text{ between 0 and x}.$$ But we know there is an $M$ so that $f''\le M$ on $[-1,1]$.
• You're saying that $M=\frac {f''(c)}{2}$ ? Jun 24 '14 at 8:37
• @GinKin: No, I'm saying that because $f''$ is continuous, it attains a maximum on a closed interval. Jun 24 '14 at 14:36
• @UneFemmeDouce: Because $f(x)\ge 0$ for all $x\in [-1,1]$ and $f(0)=0$, $x=0$ is an interior minimum point of $f$, and so $f'(0)=0$. Aug 16 '14 at 12:24