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Show that if $f$ is defined on $(a, b)$ and $c\in(a, b)$ is a local minimizer for f, then $\underline{D}f(c) \leq 0 \leq \overline{D}f(c)$.

Proof:

There exists $\delta > 0$ such that $f(c) < f(c-h)$ for $0. Then, $\frac{f(c)-f(c-h)}{h} <0$ for $0. Thus, $\underline{D}f(c) = \underline{lim}_{h\rightarrow0}\frac{f(c)-f(c-h)}{h} \leq0$.

Why can we flip f(c) and f(c-h)? Shouldn't it still be $\underline{D}f(c) = \underline{lim}_{h\rightarrow0}\frac{f(c-h)-f(c)}{h} \geq0$ which wouldn't help?

Likewise, There exists $\delta' > 0$ such that $f(c) < f(c+h)$ for $0. Then, $\frac{f(c+h)-f(c)}{h} >0$ for $0. Thus, $\overline{D}f(c) = \overline{lim}_{h\rightarrow0}\frac{f(c+h)-f(c)}{h} \geq0$.

Then, $\underline{D}f(c) \leq 0 \leq \overline{D}f(c)$.

  • 0
    Why is the first part the definition of the lower derivative?2012-12-12

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I think you meant $\frac{f(c-h)-f(c)}{-h} < 0$. But other than that, your reasoning is correct.