# Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$

let $$f$$ be a nonnegative and differentiable twice in the interval $$[-1,1]$$

Prove: if $$f(0)=0$$ and $$f'(0)=0$$ then $$f''(0)\geq 0$$

1. Are all the assumptions on $$f$$ necessary for the result to hold ?

2. what can be said if $$f''(0)= 0$$ ?

Looking at the taylor polynomial and lagrange remainder we get:

$$f(x)=f(0)+f'(0)x+\frac{f''(c)x^2}{2}$$

$$f(x)=\frac{f''(c)x^2}{2}$$

Because the function is nonnegative and $$\frac{x^2}{2}\geq 0$$ so $$f''(c)\geq 0$$

For 1., all the data is needed but I can not find a valid reason.

For 2., can we conclude that the function the null function?

Another approach is via the method of contradiction. Assume that $f''(0) < 0$. Then the function $f'$ is strictly decreasing at $0$ which means that there is a neighborhood $I$ of $0$ such that if $x \in I, x < 0$ then $f'(x) > f'(0) = 0$ and if $x \in I, x > 0$ then $f'(x) < f'(0) = 0$. We can obviously choose $I$ of the form $(-h, h)$ and hence observing sign of $f'$ in $(-h, h)$ we see that $f$ is strictly increasing in $(-h, 0]$ and strictly decreasing in $[0, h]$ and since $f(0) = 0$ it follows that $f(x) < 0$ for all $x \in I$ except $x = 0$. And this is contrary to the hypotheses that $f$ is non-negative. Thus we obtain the desired contradiction.

BTW the above argument can be replaced by the following concise argument. Since $f'(0) = 0, f''(0) < 0$ the point $0$ is a local strict maximum of $f$ and hence $f(x) < f(0) = 0$ for all $x$ in some neighborhood of $I$ of $0$ except $x = 0$. And this contradicts that $f$ is non-negative.

The argument in first paragraph actually shows how the vanishing of derivative and sign of second derivative guarantee local maxima/minima and if the reader is well aware of the proof of second derivative test for maxima/minima, it is preferable to adopt the argument in second paragraph.

As can be seen from the arguments above we don't need continuity of $f''$. And we can avoid difficult theorems like Taylor or L'Hospital. The argument used in first paragraph leads to one of the simpler proofs of Taylor's Theorem with Peano's form of remainder.

Update: The condition $f' (0)=0$ is superfluous here. The function is non-negative and $f(0)=0$ and hence $0$ is a point of local minima so that $f'(0)=0$.

Your proof is enough when $f''$ is continuous.

Here's a way without the continuity assumption.

This Taylor expansion $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + o(x^2)$ yields here: $$f(x) = \frac{x^2}{2}f''(0) + o(x^2)$$

Either $f''(0)=0$ and we're good, otherwise the previous equality rewrites as $\displaystyle \lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$

Consequently, $f(x)$ and $f''(0) x^2$ must share the same sign on a neighborhood of $0$.

Since $f\geq 0$ and $x^2\geq 0$ that implies $f''(0)\geq 0$

When $f''(0)=0$, $f(x)=o(x^2)$. I don't see anything more you could say.

• This works, so +1. You might consider referencing the Peano form of the remainder here. -Mark Jul 11, 2016 at 17:38
• How did you get to $\lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$?
– gbox
Jul 11, 2016 at 19:10
• @LeGrandDODOM it is the definition of Peano form? if we plug the limit to 2f(x) do not the limit should be $0$?
– gbox
Jul 11, 2016 at 19:25
• @LeGrandDODOM In your last comment, $f=o(g)$ if and only if $\frac{f}{g}\to 0$. Jul 11, 2016 at 22:04
• @Dr.MV Yes, typo. What I mean is $f(x) = o(g(x)) \iff \frac{f(x)}{g(x)} \to 0$ by definition and therefore $f(x) = g(x) + o(g(x)) \iff \frac{f(x)}{g(x)} \to 1$ Jul 11, 2016 at 22:18

For 2) you could have $f(x)=x^4$ which satisfies all properties and conclusions including $f''(0)=0$, yet $f(x)$ is not identically zero.

For 1) relaxing the condition that $f(0)=0$, we could look at $f(x)=\cos(x)+3$ (which has instead $f(0)=4$). It satisfies the property that $f(x)$ is non-negative, is twice differentiable on $[-1,1]$ and that $f'(0)=0$. However, $f''(0)=-\cos(0)=-1$ is not non-negative.

Relaxing the condition that $f(x)$ be non-negative on the interval, letting $f(x)=-x^2$ one has $f(0)=0, f'(0)=0$, yet $f''(0)=-2$ is not non-negative

Relaxing the condition that $f$ be twice differentiable makes no sense since then $f''(0)$ can't be talked about.

Try searching yourself as well for a function $f(x)$ which violates only the hypothesis that $f'(0)=0$ which has $f''(0)<0$.

With these counterexamples, we conclude that in fact every hypothesis was necessary.

• $x^3$ is not nonnegative on the interval. Use $x^4$ instead. Jul 11, 2016 at 17:15
• @Mosquite thanks. I moved one degree up from $x^2$ to fix that $f''(0)=0$ and forgot to account for that. Jul 11, 2016 at 17:16
• But are the hypotheses sufficient to guarantee that $f''(0)\ge 0$? Jul 11, 2016 at 17:28
• @Dr.MV I interpreted his question as asking about the after thought questions (were all in fact necessary or could any hypotheses be dropped, and can you conclude if $f''(x)=0$ that $f(x)=0$). His proof seemed sufficient to me, but LaGrandDODOM points out in his answer that the continuity of $f''(x)$ was tacitly assumed in the OP's proof, and can be dropped as well. I see no reason to replicate LaGrandDODOM's proof at this point or to come up with my own. Jul 11, 2016 at 17:32
• The OP's "proof" only shows that $f''(c)\ge 0$, for some $c\in (0,x)$, ]but does not prove that indeed $f''(0)\ge 0$. And yes, you're correct, using the Peano form for the remainder is one very straightway forward. -Mark Jul 11, 2016 at 17:36

The continuity of $f''$ is not really needed to apply Taylor's formula with the Lagrange remainder. Anyway, we may just apply l'Hopital's rule to get: $$f''(0) =\lim_{x\to 0}\frac{f'(x)-f'(0)}{x} = 2\lim_{x\to 0}\frac{f(x)}{x^2}.$$ Since both $f(x)$ and $x^2$ are non-negative over $[-1,1]\setminus\{0\}$, $\;f''(0)\geq 0$ readily follows.
Obviously we cannot deduce the strict inequality $f''(0)>0$, as shown by $f(x)=x^4$.

• @Dr.MV: I think it is better to apply L'Hospital once to get $$2\lim_{x \to 0}\frac{f(x)}{x^{2}} = \lim_{x \to 0}\frac{f'(x)}{x} = \lim_{x \to 0}\frac{f'(x) - f'(0)}{x} = f''(0) \text{ (by defn of }f''(0))$$ This works without continuity of $f''$. Jul 12, 2016 at 5:34
• @ParamanandSingh: thanks, updated. Jul 12, 2016 at 15:29
• Jul 12, 2016 at 15:46
• Thaks for link to Gowers blog post. It is indeed a very good post. BTW proof of Taylor's theorem with Peano remainder given by Gowers is essentially the same as the one I linked in my answer. Jul 13, 2016 at 6:24