Prove that there exists a point $c$ such that $f''(c)\ge 4$

Let $f:[0,1/2]\to\mathbb{R}$ be a twice differentiable function such that $f(0) = 0$ and $f'(0) = 0$. Also $f(1/2)\ge 1/2$. Prove that there exists a $c\in[0,1/2]$ such that $f''(c)\ge4$.

I have tried different approached. By Lagrange's MVT twice we can prove the that there is a $c$ such that $f''(c)\ge 2$. One of my second approaches involves taking two cases $c\le1/4$ and $c>1/4$. Another approach was to consider $g(x) = f(x) + f(x+1/4)$ or $g(x) = f(x)-f(x-1/4)$.

By Taylor's expansion $$f(x) = f(0)+\frac{f'(0)}{1!}x + \frac{f''(c_x)}{2!}x^2=\frac{f''(c_x)}{2!}x^2$$

for some $0\le c_x\le x$. Hence, at the point $x_0=1/2$ $$f(1/2)=\frac{f''(c_{x_0})}{2^3}\ge \frac{1}{2}$$ So $$f''(c_{x_0})\ge 4$$

• Nice. A useful form of the Taylor expansion. It's perhaps worth pointing out the proof of this form makes significant use of the MVT. Dec 5, 2014 at 13:26
• Maybe a stupid question, but why you stop the expansion at the second order? ok... Just saw that you have $c_x$... Dec 5, 2014 at 13:26
• @Umberto: cause we have the conditions $f(0)=f'(0)=0$ and we want to claim something about $f''(c)$.
– Mher
Dec 5, 2014 at 13:28
• @Mher thanks. I noticed what you wanted to do the moment I posted my comment... I edited it... Thanks! Dec 5, 2014 at 13:33
• How can I prove the modifier Taylor Expansion? Dec 5, 2014 at 14:57

For the original poster some more info (I hope it is useful). If $f$ is $k$ times differentiable on an open interval (in this case $]0, 1/2[$), with $f^{(k)}$ continous on the closed interval between $a$ and $x$ then the mean value form of the remained is $$R_k(x)=\frac{f^{(k)}(\xi)}{k!}(x-a)^k$$ for some real number $\xi$ between $x$ and $a$.

(taken from wikipedia to give you a quick answer, I have no books at hand at the moment: http://en.wikipedia.org/wiki/Taylor's_theorem)

We have 3 data points, $f(0) = f(0+\delta) = 0$ and $f(\tfrac{1}{2}) \geq \tfrac{1}{2}$. Using 3 points, we can try to estimate the 2nd derivative.

For any 3 data points, $a \mapsto f(a), b \mapsto f(b), c \mapsto f(c)$, the interpolating quadratic polynomial should be:

$$f(x) \approx f(a)\frac{(x-b)(b-c)(c-x)}{(a-b)(b-c)(c-a)} + f(b)\frac{(a-x)(x-c)(c-a)}{(a-b)(b-c)(c-a)} + f(c)\frac{(a-b)(b-x)(x-a)}{(a-b)(b-c)(c-a)}$$

Notice the cancellation, so we can just cross out the factors in numerator and denominator:

$$f(x) \approx \frac{(x-b)(c-x)}{(a-b)(c-a)} + \frac{(a-x)(x-c)}{(a-b)(b-c)} + \frac{(b-x)(x-a)}{(b-c)(c-a)}$$

This is just for any 3 points, there is a parabola going through it. https://stackoverflow.com/questions/717762/how-to-calculate-the-vertex-of-a-parabola-given-three-points

In our case: $a=0,b = 0 + \delta, c = \tfrac{1}{2}$:

$$f(x) \approx 0\cdot \frac{(x-\delta)(\tfrac{1}{2}-x)}{(-\delta )\tfrac{1}{2}} + 0\cdot \frac{(-x)(x-\tfrac{1}{2})}{(-\delta)(\delta-\tfrac{1}{2})} + (\tfrac{1}{2} + \epsilon) \cdot \frac{(\delta-x)x}{(\delta-\tfrac{1}{2})(\tfrac{1}{2})} \approx \boxed{(2+\epsilon)x^2 }$$

with $\epsilon > 0$. In this case the 2nd derivative should be something like: $f''(x) \approx 4 + \epsilon$.

The Mean Value Theorem should tell us this point with $f''(c) = 4$ is a topological feature and can't be removed.

See section 4 of Osculating curves: around the Tait-Kneser Theorem by Ghys, Tabachnikov and Timorin.