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Hi everyone I have some trouble with one point in the following proof.

Let $f$ be a convex function (strict convex function) on a real interval. If $f'_-(a)=f'_+(a)$ where $f'_-$ and $f'_+$ are the left-hand and right-hand derivative respectively, then $f'_+$ is continuous at $a$.

Proof: Given $\epsilon>0$, there is a $\delta>0$ such that

$$0<x-a<\delta\implies \left|\frac{f(x)-f(a)}{x-a}-f'_+(a)\right|<\epsilon/2 $$

Since the quotients are always bigger than the right derivative, the difference is always non-negative. The right-hand derivative implies that $f$ is continuous to the right, and so we can choose some $b>a$ for which

$$\frac{f(x)-f(b)}{x-b}<\frac{f(x)-f(a)}{x-a}+\epsilon/2\tag{*}$$

and thus

$$\begin{align}f'_+(b)<\frac{f(x)-f(b)}{x-b}<\frac{f(x)-f(a)}{x-a}+\epsilon/2 \\ <f'_+(a)+\epsilon\end{align}$$

Since $f'_+$ is increasing because $f$ is convex, so for $\,a<x\le b$ we must have

$$f'_+(a)<f'_+(x)\le f'_+(b)<f'_+(a) +\epsilon $$ $$0< f'_+(x)-f'_+(a) < \epsilon$$

which shows that $\lim_{x\downarrow a}f'_+(x)=f'_+(a)$ a similar argument shows that $\lim_{x\uparrow a}f'_+(x)=f'_+(a)$.

So my question is regarding to (*). I understand that is intuitively clear that always we can pick such $b$, by the right continuity of $f$, but I have failed to give a formal $\epsilon$-$\delta$ argument. I'd appreciate any help, thanks.

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for a given $c$,$\quad$ $0<c-a< \delta$,$\quad$ $\Rightarrow$ $\quad$ $|\frac{f(c)-f(a)}{c-a}-f_{+}^{'}(a)|<\epsilon/2$.

$f(x)$ is right hand continuous at $x=a$, so is $g(x)=\frac{f(c)-f(x)}{c-x}$.

for every $\quad\epsilon >0 \quad$ there's some $\quad\delta^{'}>0 \quad$ such that if $\quad$$0<x<\delta^{'}, \quad$then:

$|g(x)-g(a)|<\epsilon\quad \Rightarrow\quad |\frac{f(c)-f(x)}{c-x}-\frac{f(c)-f(a)}{c-a}|<\epsilon.\quad$

so for every $x,\quad a<x<min(\delta^{'},c-a),\quad |\frac{f(c)-f(x)}{c-x}-\frac{f(c)-f(a)}{c-a}|<\epsilon/2.\quad \Rightarrow \quad \frac{f(c)-f(x)}{c-x}<\frac{f(c)-f(a)}{c-a}+\epsilon/2$

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