# Infimum of right derivative and infimum of left derivative are equal?

Suppose we have continuous function $$f: [a,b] \to \mathbb{R}$$ with right and left derivatives on $$(a,b)$$. So would $$m_-=\inf\{f_+'(x):x \in (a,b)\}= \inf\{f_-'(x):x \in (a,b)\}=m_+$$ take place?

I tried to prove by contradiction. Suppose $$m_-. Then exists $$x_0 \in (a,b)$$, such that $$f'_-(x_0)<0.5(m_++m_-)$$.

Then if we take $$\epsilon=0.25(m_+-m_-)$$, then exists $$\delta > 0$$ such that if $$0 then $$|\frac{f(x)-f(x_0)}{x-x_0}-f'_-(x_0)|<\epsilon$$.

Then I choose $$y \in (x_0-\delta; x_0)$$. We have $$\frac{f(y)-f(x_0)}{y-x_0}<\epsilon+ f'_-(x_0)<0.75m_++0.25m_-$$

Then if $$x_0 \to y+0$$, it appears that $$f'_+(y), which is contradiction.

I don't know if it's correct, so I would be glad if you point me in the right direction.

• You don't have any control over $x_0$ and you cannot let $x_0 \to y$. Sep 28 '20 at 8:44
• It is not appropriate to delete a question after you have gotten an answer. This is disrespectful to those who have taken the time to help you out. Please do not do this in the future. Sep 28 '20 at 15:28

As pointed out in the comments, your proof does not work because $$x_0$$ is chosen first and $$y$$ depends on $$x_0$$, therefore the limiting process $$x_0 \to y$$ is not valid.
For a valid proof you can use that a function with non-negative right derivative at each point is non-decreasing (see for example $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing).
Then you can argue as follows: Assume that $$m=\inf\{f_+'(x):x \in (a,b)\}$$ is finite, and set $$g(x) = f(x) - mx$$. Then $$g_+'(x) \ge 0$$ on $$(a, b)$$, so that $$g$$ is non-decreasing. It follows that $$g_-'(x) \ge 0$$ and therefore $$f_-'(x) \ge m$$ on $$(a, b)$$.
This proves that $$\inf\{f_+'(x):x \in (a,b)\} \le \inf\{f_-'(x):x \in (a,b)\}$$ and the reverse inequality can be proved in the same way, or by considering $$\tilde f(x) = -f(a+b-x)$$ (so that a right derivative of $$f$$ becomes a left derivative of $$\tilde f$$, and vice versa).
For a self-contained proof one can argue as follows: Again assume that $$m=\inf\{f_+'(x):x \in (a,b)\}$$ is finite. For $$a < c < d < b$$ consider the function $$g(x) = f(x) - \frac{f(d)-f(c)}{d-c}(x-c) \, .$$ Then $$g(c) = g(d)$$ so that the maximum of $$g$$ is attained at some point $$x_0 \in [c, d)$$. Then $$0 \ge g_+'(x_0) = f_+'(x_0) - \frac{f(d)-f(c)}{d-c} \ge m - \frac{f(d)-f(c)}{d-c} \\ \implies \frac{f(d)-f(c)}{d-c} \ge m \, .$$ It follows that $$f_-'(d) \ge 0$$. This holds for all $$d \in (a, b)$$, so that $$\inf\{f_-'(x):x \in (a,b)\} \ge m = \inf\{f_+'(x):x \in (a,b)\} \, .$$ As before, the reverse inequality can be proved in the same way or by symmetry arguments.