# Twice differentiable continuous function estimate

• Dec 3rd 2008, 05:57 AM
Twice differentiable continuous function estimate
Suppose that a function $f: \mathbb {R} \rightarrow \mathbb {R}$ is twice differentiable and both $f', f''$ are continuous, also suppose that $f(0)=f'(0)=0$, show that $\exists c>0$ such that $|f(x)| \leq cx^2$.

Proof so far.

By the Taylor's Theorem, we know that $f(x) = f(0)+f'(0)x+ \frac {1}{2} f''( \epsilon ) x^2$ where $\epsilon$ is a number between 0 and x.

So we have $f(x) = \frac {1}{2} f''( \epsilon ) x^2$

Let $c= \frac {1}{2}f''( \epsilon )$, then $|f(x)| \leq cx^2$.

Is this right? Thanks.
• Dec 3rd 2008, 08:16 AM
Opalg
There's something wrong with this question, if the inequality $|f(x)|\leqslant cx^2$ is supposed to hold for all $x\in\mathbb{R}$. For example, the function $f(x)=x^4$ satisfies the hypotheses of the question but not the conclusion.

The best you can realistically ask for is that there exists some neighbourhood of the origin, say $|x|\leqslant a$, such that $|f(x)|\leqslant cx^2$ whenever $|x|\leqslant a$.

With that amendment to the question, your proof is partially correct. But there's a defect in it. The problem is that your choice of $\epsilon$ depends on x. You need to adjust the proof so as to get an estimate of the size of $f''(\epsilon)$ that does not depend on x. This is possible because you are told that f'' is continuous. That means that it is bounded in any closed bounded interval. So given the interval $-a\leqslant x\leqslant a$ there exists M>0 such that $|f''(x)|\leqslant M$ for all x in that interval. You can then use M in place of $f''(\epsilon)$ in your proof, define $c=\tfrac12M$, and the proof will work nicely.
• Dec 3rd 2008, 09:35 AM
Thanks for the help.

The second part of this question is to find an exmaple of a differentiable function $f: \mathbb {R} \rightarrow \mathbb {R}$ such that $f(0)=f'(0)=0$, but yet no inequality of the form $|f(x)| \leq cx^2$ holds in any neighborhood of the origin.

So I guess $f(x) = x^4$ would be the example then?
• Dec 3rd 2008, 10:16 AM
Opalg
Quote:

Thanks for the help.

The second part of this question is to find an exmaple of a differentiable function $f: \mathbb {R} \rightarrow \mathbb {R}$ such that $f(0)=f'(0)=0$, but yet no inequality of the form $|f(x)| \leq cx^2$ holds in any neighborhood of the origin.

So I guess $f(x) = x^4$ would be the example then?

No! – because $f(x) = x^4$ does satisfy an inequality of that form in a neighbourhood of the origin. It just doesn't satisfy such an inequality globally, which is what you were previously asking for.

To get the right sort of example here, you need to ask how the proof of the first part of the question could go wrong. What you should notice is that it depended crucially on f''(x) being bounded in a neighbourhood of the origin. So you need to look for a candidate for f''(x) that is unbounded near the origin. But it must have a continuous integral f'(x). The sort of function that might do the job is $f''(x) = 1/\sqrt x$. Try integrating that twice and see if it gives a function f(x) with the required properties.
• Dec 4th 2008, 06:14 AM
Claim: $f(x) = \frac {4}{3}x^{ \frac {3}{2} }$ satisfies the hypothesises but the inequality $|f(x)| \leq cx^2$ do not hold in any neighborhood of the origin.

Proof so far.

Now, $f'(x)=2x^{ \frac {1}{2} }$, so $f(0)=f'(0)=0$

I need to show that in any neighborhood of the origin, $|f(x) > cx^2 \ \ \ \ \ \forall c>0$

Given $c>0$ and $|x| \leq a$ for some $a \in \mathbb {R}$, we have $|f(x)|=|\frac {4}{3} x^ { \frac {3}{2} } | = \frac {4}{3} |x| ^ { \frac {3}{2} } \geq \frac {4}{3} (-a)^{ \frac {3}{2} }$

But how would I get a positive number on the right hand side? Thanks.
• Dec 4th 2008, 08:18 AM
Opalg
Quote:

Claim: $f(x) = \frac {4}{3}x^{ \frac {3}{2} }$ satisfies the hypothesises but the inequality $|f(x)| \leq cx^2$ do not hold in any neighborhood of the origin.
Now, $f'(x)=2x^{ \frac {1}{2} }$, so $f(0)=f'(0)=0$
I need to show that in any neighborhood of the origin, $|f(x) > cx^2 \ \ \ \ \ \forall c>0$
Given $c>0$ and $|x| \leq a$ for some $a \in \mathbb {R}$, we have $|f(x)|=|\frac {4}{3} x^ { \frac {3}{2} } | = \frac {4}{3} |x| ^ { \frac {3}{2} } \geq \frac {4}{3} (-a)^{ \frac {3}{2} }$
Do this by contradiction. Suppose that there is a neighbourhood U of 0, and a constant c>0, such that $\left|\tfrac43x^{3/2}\right|\leqslant cx^2$ for all x in U. Then in particular that would hold for all positive x in U, so that $\tfrac43x^{3/2}\leqslant cx^2$ whenever $0. Square both sides, divide by x^3, and you see that $x\geqslant \frac{16}{9c^2}$. But since U is a neighbourhood of 0 it must certainly contain points x satisfying $0. That contradiction shows that no such neighbourhood U and constant c can exist.