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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 $m_-<m_+$. 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 $0<x_0-x<\delta$ 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 $f'_+(y)<m_+$, which is contradiction.

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

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  • $\begingroup$ You don't have any control over $x_0$ and you cannot let $x_0 \to y$. $\endgroup$ Sep 28 '20 at 8:44
  • $\begingroup$ 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. $\endgroup$
    – Xander Henderson
    Sep 28 '20 at 15:28
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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.

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