Right versus left derivative by increasing one-sided derivatives

Question

Condider a map $f:D\to\mathbb{R}$, $D$ open interval. Let be $y,z\in D$, $y<z$. Suppose that $f$ has everywhere in $D$ both left and right derivative (this implies that $f$ is continuous), both increasing. I want to show that $$ f'(y^+)\le f'(z^-). $$

I have found a way, but I'm not completely convinced that it is formally correct: $$ f'(y^+) =\lim_{\varepsilon\to 0^+}\frac{f(y+\varepsilon)-f(y)}{\varepsilon} \overset{\forall\delta}{\le} f'((z-\delta)^+) = \lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon} \overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon} =f'(z^-) $$ where $\delta>0$ and $y<z-\delta$.

Is there a way to formalise my proof?

Answer 1

(I'll use the notation $f'_+$ and $f'_-$ for the right and left derivative.)

Unfortunately your proof does not work. This $$ \lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon} \overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon} $$ makes no sense because $\delta$ is fixed and the limits are taken for $\varepsilon \to 0 $. Also it would imply that $f'_+(z-\delta) = f'_-(z)$ for $0 < \delta < z-y$, which is true only if $f$ is linear between $y$ and $z$.

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What we can show is that $$ \tag{*} f'_+(y) \le \frac{f(z)-f(y)}{z-y} \le f'_-(z) $$ for $y < z$, which implies the desired conclusion. (This is motivated by the fact that both an increasing right derivative and an increasing left derivative imply that $f$ is convex.)

The proof of $(*)$ mimics the proof of Rolle's theorem and the mean-value theorem. We consider the function $$ g(x) = f(x) - (x-y)\frac{f(z)-f(y)}{z-y} $$ which is continuous and satisfies $g(y) = g(z)$. It follows that $g$ attains its maximum on the interval $[y, z]$ at some point $w \in [y, z)$. Then $g'_+(w) \le 0$ and it follows that $$ f'_+(y) \le f'_+(w) = g'_+(w) + \frac{f(z)-f(y)}{z-y} \le \frac{f(z)-f(y)}{z-y} \, . $$ This proves the left inequality in $(*)$, the proof of the right inequality works similarly.

Answer 2

We can prove the stronger statement $$\lim_{x\to y^+}f'_-(x)=f'_+(y)$$

By the Monotone Convergence Theorem $\lim_{x\to y^+}f'_-(x)$ exists. Call that limit $A$.

Now, let $\epsilon>0$ be given and choose a value of $\delta>0$ such that both of these are true:
$$x\in(y,y+\delta] \implies f'_-(x)\in \left[A,A+\frac{\epsilon}2 \right]$$ and $$\left|\frac{f(y+\delta)-f(y)}{\delta}-f'_+(y)\right|\le \frac{\epsilon}2$$ That is possible because of the definitions of the left hand derivative and the limit.

Similar to the answer from @MartinR, define the function $g(x)=f(x)-\frac{f(y+\delta)-f(y)}{\delta}(x-y)$ so that $g(y)=g(y+\delta)$ and both $g'_-(x)=f'_-(x)-\frac{f(y+\delta)-f(y)}{\delta}$ and $g'_+(x)=f'_+(x)-\frac{f(y+\delta)-f(y)}{\delta}$ are increasing.

Claim 1. $g'_+(y) \le 0$
This follows because if it were not true, then $g'_+(x)$ would have to be positive for all $x\in (y,y+\delta)$. That would imply $g$ is increasing (proof here) in that interval and it would be impossible for $g(y+\delta)$ to equal $g(y)$.

Claim 2. $g'_-(y+\delta) \ge 0$
Same argument as Claim 1. If it were not true, then $g'_-$ would have to be negative in the whole interval which would make it impossible for the function to have the same value at both ends of the interval.

Now,
$$g'_-(y+\delta)\ge0$$ $$f'_-(y+\delta)-\frac{f(y+\delta)-f(y)}{\delta}\ge0$$ $$f'_-(y+\delta) \ge \frac{f(y+\delta)-f(y)}{\delta} \ge f'_+(y)-\frac{\epsilon}2$$ We also know that $f'_-(y+\delta)\in \left[A,A+\frac{\epsilon}2 \right]$

Thus, $$A \ge f'_-(y+\delta)-\frac{\epsilon}2 \ge f'_+(y)-\epsilon \label{eq1}\tag{1}$$

On the other hand,
$$g'_-(y)\le0$$ $$f'_-(y)-\frac{f(y+\delta)-f(y)}{\delta}\le0$$ $$f'_-(y) \le \frac{f(y+\delta)-f(y)}{\delta} \le f'_+(y)+\frac{\epsilon}2$$

So, $$A \le f'_-(y+\delta) \le f'_+(y)+\frac{\epsilon}2 \label{eq2}\tag{2}$$

Combining the two inequalities (1) and (2), since $\epsilon$ was arbitrary, we have $A=f'_+(y)$.

To answer the original question, if $z>y$ then by the monotonicity assumption $f'_-(z)>\lim_{x\to y^+}f'_-(x)=f'_+(y)$