Mathematics Asked by Wyatt on February 22, 2021

Condider a map $f:Dtomathbb{R}$, $D$ open interval.

Let be $y,zin D$, $y<z$. 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_{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 $y<z-delta$.

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

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