# a problem of existence of local maxima and minima

Let $k:[0,1]→\mathbb{R}$ be a continuous function. Suppose that $k$ has local maximum at two distinct points $x_1 , x_2$ in $[0,1]$. show that $k$ has a local minimum at some point $x_3$ in [$0,1$].

I am totally stuck on it.can anyone help me please to solve the above problem.thanks for your time.

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A continuous function on a compact set attains a minimum. – nullUser Mar 13 '13 at 14:59
then what is the use of the fact given "k has local maximum at two distinct points $x_1,x_2$ in [$0,1$]" – priti Mar 13 '13 at 15:06

## 1 Answer

We use the basic fact that a function continuous on a closed interval attains an absolute maximum and an absolute minimum in that interval.

We show that under our conditions there is a point $c$ in the open interval $(x_1,x_2)$ at which our function (which I will call $f$) attains an absolute minimum, and therefore a local minimum.

There is only one slightly tricky point on which one could stumble: it is possible that $f$ attains its absolute minimum at one of $x_1$ or $x_2$. The two cases are dealt with similarly.

So suppose that $f$ attains an absolute minimum for the interval $[x_1,x_2]$ at $x_1$. Because we have a local maximum at $x_1$, there is a interval $[x_1,x_1+\epsilon)$ such that $f(x_1)\ge f(x)$ for all $x$ in $[x_1,x_1+\epsilon)$. But because at $x_1$ we have an absolute minimum for the interval $[x_1,x_2]$, we also have $f(x_1)\le f(x)$ for all $x$ in the interval $[x_1,x_1+\epsilon)$.

It follows that $f(x)=f(x_1)$ for all $x$ in $[x_1,x_1+\epsilon)$. Thus $f$ attains a local minimum at, for example, $x_1+\epsilon/2$.

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