2
$\begingroup$

Full Problem Statement: Let $f'(x_0)>0$ for some point $x_0$ in the interior of the domain $D$ of $f$ show that there is a $\delta>0$ so that $f(x)<f(x_0)<f(y)$ whenever $x_0-\delta<x<x_0<y<x_0+\delta.$

My Attempt: Since $x_0$ is an interior point of $I$ there exists a ball completely contained inside the interval. Let radius of that ball be $r$ and so we have that $(x_0-r,x_0+r)\subseteq D.$ Then if $x_0-r<x<x_0<y<x_o+r$ and $f(x_0)\geq f(y)$ then we have that $$f'(x_0)=f_{+}'(x_0)=\lim_{y\to x_0}\frac{f(y)-f(x_0)}{y-x_0}\leq 0$$ which is a contradiction. Similarily we can show that $f(x)<f(x_0).$ I am not sure whether this argument is correct and so would appreciate any insight.

$\endgroup$
2
$\begingroup$

No, the $y$ for which $f(x_{0})\geq f(y)$ is a particular $y$, but the $y$ in the limit $\lim_{y\rightarrow x_{0}}$ are all sufficiently closed $y$ to $x_{0}$.

Actually you can prove that statement in direct way. Find some $\delta>0$ such that for all $x\in(x_{0}-\delta,x_{0}+\delta)-\{x_{0}\}$ \begin{align*} \left|\dfrac{f(x)-f(x_{0})}{x-x_{0}}-f'(x_{0})\right|<\dfrac{1}{2}f'(x_{0}), \end{align*} expand the inequality get \begin{align*} \dfrac{f(x_{0})-f(x)}{x_{0}-x}-f'(x_{0})>-\dfrac{1}{2}f'(x_{0}),~~~~x\in(x_{0}-\delta,x_{0}), \end{align*} so \begin{align*} f(x_{0})-f(x)>\dfrac{1}{2}f'(x_{0})(x_{0}-x)>0,~~~~x\in(x_{0}-\delta,x_{0}), \end{align*} andn similarly, \begin{align*} f(y)-f(x_{0})>\dfrac{1}{2}f'(x_{0})(y-x_{0})>0,~~~~y\in(x_{0},x_{0}+\delta). \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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