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Show that the function is not differentiable at the point z

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,154
Hi!! Could you help me with the following?
Let [tex] g: R \rightarrow R [/tex] a bounded function. There is a point [tex] z \epsilon R [/tex] for which the function [tex] h: R [/tex] \ [tex]\{z\} \rightarrow R [/tex], where [tex] h(x)=\frac{g(x)-g(z)}{x-z} [/tex] is not bounded. Show that the function [tex] g [/tex] is not differentiable at the point [tex] z [/tex].

My idea is the following:
Let the function [tex] g [/tex] be differentiable at the point [tex] z [/tex], so [tex] lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L [/tex]. So there is a ε>0: [tex] |\frac{g(x)-g(z)}{x-z}-L|<ε [/tex].
[tex] |h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L| [/tex].
So the function [tex] h[/tex] is bounded. That cannot be true. So the function [tex] g [/tex] is not differentiable at the point [tex] z [/tex].

Is my idea right??
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi!! Could you help me with the following?
Let [tex] g: R \rightarrow R [/tex] a bounded function. There is a point [tex] z \epsilon R [/tex] for which the function [tex] h: R [/tex] \ [tex]\{z\} \rightarrow R [/tex], where [tex] h(x)=\frac{g(x)-g(z)}{x-z} [/tex] is not bounded. Show that the function [tex] g [/tex] is not differentiable at the point [tex] z [/tex].

My idea is the following:
Let the function [tex] g [/tex] be differentiable at the point [tex] z [/tex], so [tex] lim_{x \rightarrow z}{ \frac{g(x)-g(z)}{x-z}}=L [/tex]. So there is a ε>0: [tex] |\frac{g(x)-g(z)}{x-z}-L|<ε [/tex].
[tex] |h(x)|=|h(x)-L+L| \leq |h(x)-L|+|L|=|\frac{g(x)-g(z)}{x-z}-L|+|L|<ε+|L| [/tex].
So the function [tex] h[/tex] is bounded. That cannot be true. So the function [tex] g [/tex] is not differentiable at the point [tex] z [/tex].

Is my idea right??
Absolutely correct! (Sun)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,154
Great!!! (Clapping)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,154
Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Now I'm looking again the exercise..
There is a $ε>0$ such that $|\frac{g(x)-g(z)}{x-z}-L|<ε$ for $|x-z|<δ$, right? So $h$ is bounded for $x$ such that $|x-z|<δ$.
Do I have to show also that $h$ is bounded for $x$ such that $|x-z|>δ$, or is the above enough?
Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,154
Yes, you need to show that if $h$ is unbounded, then it must be unbounded near $z$, not anywhere else. But if $|x-z|\geqslant \delta$ then $\left|\dfrac1{x-z}\right|$ is bounded by $1/\delta$. Also, you know that $g$ is bounded. So it follows that $h$ must indeed be bounded outside the region $|x-z|< \delta$.
So do I have to show both cases $ |x-z|< \delta$ and $ |x-z|\geq \delta$ ??