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...

Please help, thank you!


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 $[-1,1]$.

  • $\begingroup$ Great! thanks a lot! $\endgroup$
    – Galc127
    Jun 22 '14 at 11:30
  • $\begingroup$ You're saying that $M=\frac {f''(c)}{2}$ ? $\endgroup$
    – GinKin
    Jun 24 '14 at 8:37
  • $\begingroup$ @GinKin: No, I'm saying that because $f''$ is continuous, it attains a maximum on a closed interval. $\endgroup$ Jun 24 '14 at 14:36
  • $\begingroup$ Right. From extreme value theorem. $\endgroup$
    – GinKin
    Jun 24 '14 at 14:39
  • 1
    $\begingroup$ @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 $f'(0)=0$. $\endgroup$ Aug 16 '14 at 12:24

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