If $f$ is monotone increasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0}) \geq 0$.

alternative-proofderivativesreal-analysissolution-verification

Let $X$ be a subset of $\textbf{R}$, let $x_{0}\in X$ be a limit point of $X$, and let $f:X\rightarrow\textbf{R}$ be a function. If $f$ is monotone increasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0}) \geq 0$. If $f$ is monotone decreasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0})\leq 0$.

MY ATTEMPT

Lemma

Let $X\subseteq\textbf{R}$, $f:X\rightarrow\textbf{R}$, $g:X\rightarrow\textbf{R}$, $x_{0}\in X$ is an adherent point, $f(x) \leq g(x)$ for every $x\in X$ and $\displaystyle\lim_{x\rightarrow x_{0}}f(x) = L$ and $\displaystyle\lim_{x\rightarrow x_{0}}g(x) = M$. Then we have that $L \leq M$.

Proof

According to the definition of limit, for every $\varepsilon > 0$, there are $\delta_{1} > 0$ and $\delta_{2} > 0$ such that
\begin{align*}
\begin{cases}
0 < |x – x_{0}| < \delta_{1}\\\\
0 < |x – x_{0}| < \delta_{2}
\end{cases} \Longrightarrow
\begin{cases}
|f(x) – L| < \varepsilon\\\\
|g(x) – M| < \varepsilon
\end{cases} \Longrightarrow L – \varepsilon < f(x) \leq g(x) < M + \varepsilon
\end{align*}

Let us assume that $L > M$. In this case, we can choose $\displaystyle\varepsilon = \frac{L – M}{3}$, whence we get that
\begin{align*}
M – L + 2\varepsilon > M – L + \frac{2(L – M)}{3} = \frac{M – L}{3} > 0 \Longrightarrow M > L
\end{align*}
which leads to a contradiction. Therefore the original claim is true and $L \leq M$.

Solution

Assuming that $f$ is monotone increasing at $x_{0}$, we have that
\begin{align*}
\frac{f(x) – f(x_{0})}{x – x_{0}} \geq 0
\end{align*}

Taking the limit from both sides to $x_{0}$ we conclude that
\begin{align*}
\lim_{x\rightarrow x_{0}}\frac{f(x) – f(x_{0})}{x – x_{0}} = f'(x_{0}) \geq 0 = \lim_{x\rightarrow x_{0}}0
\end{align*}

simliar reasoning applies to the monotone decreasing case, and we are done.

Could someone please verify if I am arguing correctly? Any other solution is welcome.

Best Answer

Why did you even write the lemme? It looks like your solution is self-contained and correct. Perhabs this helps you : $$(D_{x_0}f)(x):=\frac{f(x)-f(x_0)}{x-x_0}$$ is continuous everywhere, by the definition of the $f$. Then by your argument, ($D_{x_0}f\geq 0$ everywhere) $$\lim_{x\rightarrow x_0}(D_{x_0}f)(x)\geq 0.$$

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