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Got this from my Real Analysis problem set:

Suppose $f(x)$ continuous on some open interval $I$, and $c$ is maximum point for $f(x)$ inside this interval. Is it true that that $f(x)$ is increasing immediately in the left of $c$ and immediately decreasing in the right of $c$?

(The constant function is not a counterexample because it's considered to both increasing and decreasing.)

I really think so. Am I wrong?

If the function strictly incraeses after $c$, so $c$ cannot be a maximum.

Am I not seeing something important. This seems obvious :/

Any help would be awesome, thanks in advance!

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Consider the function $x\sin\frac1x-2|x|$

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    $\begingroup$ It's actually easier with your original answer with $-|x|$, since, when $x\sin{1/x}=|x|$ for inifnitely many points near $0$, but it is not identically $0$. $\endgroup$ – Thomas Andrews Jan 28 '16 at 15:44
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What you say makes some sense but this is not really what is meant. What is meant is:

Is there some $\epsilon > 0 $ such that on such $f$ is a decreasing function on $[c, c+\epsilon]$, that is in this interval $f(x) \ge f(y)$ whenever $x \le y $. (While you seem to read it as $f(c) \ge f(x)$ for all $x$.)

This is somewhat counter-intuitively not true. A counter-example is sketched in another answer. The point is to take something that oscillates very fast near $c$.

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