# Right versus left derivative by increasing one-sided derivatives

Mathematics Asked by Wyatt on February 22, 2021

Condider a map $$f:Dtomathbb{R}$$, $$D$$ open interval.
Let be $$y,zin D$$, $$y. Suppose that $$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:
$$f'(y^+) =lim_{varepsilonto 0^+}frac{f(y+varepsilon)-f(y)}{varepsilon} overset{foralldelta}{le} f'((z-delta)^+) = lim_{varepsilonto 0^+}frac{f(z-delta+varepsilon)-f(z-delta)}{varepsilon} overset{varepsilon=delta}{=} lim_{varepsilonto 0^+}frac{f(z)-f(z-varepsilon)}{varepsilon} =f'(z^-)$$
where $$delta>0$$ and $$y.

Is there a way to formalise my proof?

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

Unfortunately your proof does not work. This $$lim_{varepsilonto 0^+}frac{f(z-delta+varepsilon)-f(z-delta)}{varepsilon} overset{varepsilon=delta}{=} lim_{varepsilonto 0^+}frac{f(z)-f(z-varepsilon)}{varepsilon}$$ 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 $$z$$.

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.

Correct answer by Martin R on February 22, 2021

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

By the Monotone Convergence Theorem $$lim_{xto y^+}f'_-(x)$$ 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:
$$xin(y,y+delta] implies f'_-(x)in left[A,A+frac{epsilon}2 right]$$ 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 $$xin (y,y+delta)$$. 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 $$f'_-(z)>lim_{xto y^+}f'_-(x)=f'_+(y)$$

Answered by John L on February 22, 2021