Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been studying "Berkeley Problems in Mathematics, Souza, Silva" and I came across this problem:

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^{\infty}$ function. Assume that $f(x)$ has a local minimum at $x = 0$. Prove there is a disc centered on the $y$ axis which lies above the graph of $f$ and touches the graph at $(0, f(0))$.

We use Taylor's theorem:

there is a constant $C$ such that $|f(x) - f(0) - f’(0)x| \le Cx^2$ and we assume that $|x| < 1$.

Why is that?

I know that if a function has a local minimum at $0$, it means that in a certain neighbourhood its values cannot be less than $f(0)$.

Will anything bad happen if we instead assume that $|x|<\delta<1$ ?

Please help me. I see it's a crucial step in the solution of this problem.

http://thor.info.uaic.ro/~fliacob/An1/2012-2013/Concursuri/SEEMOUS-2013/Baza%20de%20documentare/Souza,%20Silva%20-%20Berkeley%20Problems%20In%20Mathematics%20(440S).pdf

question: Problem 1.4.26 page 24 , solution: page 177

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Suppose that $|f''(x)|\le M$ for $|x|\le\epsilon$. Since $f'(0)=0$, we have that for $|x|\lt\epsilon$, $$ \left|\frac{f'(x)-f'(0)}{x-0}\right|=|f''(\xi)|\le M $$ for some $\xi$ between $0$ and $x$; that is, for $|\xi|\le\epsilon$.

Thus, $|f'(x)|\le M|x|$ and integrating yields $|f(x)-f(0)|\le\frac M2x^2$ for $|x|\le\epsilon$.

The bottom of the circle of radius $r$ centered at $f(0)+r$ is $$ g(x)=f(0)+r-\sqrt{r^2-x^2} $$ so that $g(0)=f(0)$ and $$ \begin{align} g(x)-f(0) &=\frac{x^2}{r+\sqrt{r^2-x^2}}\\ &\ge\frac1{2r}x^2 \end{align} $$ Thus, if we set $r\le\min\left(\frac1M,\epsilon\right)$, the circle of radius $r$ centered at $f(0)+r$ satisfies the requirements. That is, $$ g(x)\ge f(0)+\frac1{2r}x^2\ge f(0)+\frac M2x^2\ge f(x)\\ g(0)=f(0) $$ and the circle only extends to $|x|\le r\le\epsilon$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.