# Prove this derivative is continuous

• Dec 6th 2008, 06:20 PM
Prove this derivative is continuous
Let $\displaystyle U$ be a neighborhood of 0 in $\displaystyle \mathbb {R}$.

Suppose that $\displaystyle f:U \rightarrow \mathbb {R}$ is twice differentiable , f' and f'' are continuous, and $\displaystyle f(0)=0$.

Define $\displaystyle g: U \rightarrow \mathbb {R}$ by $\displaystyle g(x)= \frac {f(x)}{x}$ if $\displaystyle x \neq 0$ and $\displaystyle \lim _{ h \rightarrow 0 } \frac {f(h)}{h}$ if $\displaystyle x = 0$

Show that g' exists and is continuous on U.

Proof so far.

Claim: g' exists on U.

By the division theorem, we know that $\displaystyle f(x)=xg(x)$, implies that for $\displaystyle x \neq 0$, we would have $\displaystyle f'(x)=xg'(x)+g(x)$, so we have $\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}$.

For $\displaystyle x=0$, we have $\displaystyle g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0$

Claim: g' is continuous on U.

Write $\displaystyle f(x)=f(0)+f'(0)x+f'(t)x^2 \ \ \ \ \ 0<t<x$, so we have $\displaystyle f(x)=f'(0)x+f''(xh)x^2 \ \ \ \ \ 0<h<1$

In the case that $\displaystyle x \neq 0$:

Given $\displaystyle \epsilon > 0$, pick $\displaystyle \delta > 0$, then for each $\displaystyle x \in U$ and $\displaystyle |x-x_0|< \delta$, we have $\displaystyle |g'(x)-g'(x_0)|=| \frac {f'(x)- \frac {f(x)}{x}}{x} - \frac {f'(x_0)- \frac {f(x_0)}{x}}{x}|$$\displaystyle = \frac {|f'(x)-f'(x_0)|}{x} + \frac {| \frac {f(x)}{x}- \frac {f(x_0)}{x_0}|}{x} I have a feeling I'm very wrong as I'm stuck at this point, any hints? Thank you. • Dec 7th 2008, 04:22 AM Laurent Quote: Originally Posted by tttcomrader Let \displaystyle U be a neighborhood of 0 in \displaystyle \mathbb {R} . Suppose that \displaystyle f:U \rightarrow \mathbb {R} is twice differentiable , f' and f'' are continuous, and \displaystyle f(0)=0. Define \displaystyle g: U \rightarrow \mathbb {R} by \displaystyle g(x)= \frac {f(x)}{x} if \displaystyle x \neq 0 and \displaystyle \lim _{ h \rightarrow 0 } \frac {f(h)}{h} if \displaystyle x = 0 Show that g' exists and is continuous on U. Proof so far. Claim: g' exists on U. By the division theorem, we know that \displaystyle f(x)=xg(x), implies that for \displaystyle x \neq 0 , we would have \displaystyle f'(x)=xg'(x)+g(x), so we have \displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} . For \displaystyle x=0, we have \displaystyle g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0 - Note that in fact \displaystyle g(0)=f'(0) (the limit defining \displaystyle g(0) exists because \displaystyle f is differentiable at 0). - Your last limit is incorrect. You can't just replace \displaystyle g(0) by \displaystyle \frac{f(h)}{h}. You could write instead: for any \displaystyle h\neq 0 in \displaystyle U, \displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}, and use Taylor's theorem to show that the previous quantity converges to \displaystyle \frac{f''(0)}{2}. Quote: Claim: g' is continuous on U. First, you wrote that \displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} , so it should be clear (from the hypotheses) that \displaystyle g' is continuous on \displaystyle U\setminus\{0\}. The only thing to check is that \displaystyle \lim_{h\to 0}g'(h)=g'(0) (\displaystyle =\frac{f''(0)}{2} because of the previous question) . For any \displaystyle h\neq 0 (and \displaystyle h\in U), \displaystyle g'(h)=\frac{hf'(h)-f(h)}{h^2}, so you have to apply Taylor's theorem to \displaystyle f and to \displaystyle f'. • Dec 7th 2008, 06:54 AM tttcomrader My work so far: 1. Use the Taylor's Theorem to show that \displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2} converges to \displaystyle \frac{f''(0)}{2} Proof. By the Taylor's Theorem, write \displaystyle f(x)=f(0)+f'(0)x+f''(t)x^2 \ \ \ \ \ 0<t<x , since \displaystyle f(0)=0, we have \displaystyle f(x)=f'(0)x+f''(t)x^2, implies that \displaystyle f(h)=f'(0)h+f''(t)h^2 \ \ \ \ \ 0<t<h. Then \displaystyle g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf''(0)}{h^2} \displaystyle = \lim _{h \rightarrow 0} \frac {f'(0)h+f''(t)h^2-f''(0)h}{h^2}= \lim _{h \rightarrow 0}f''(t) Now, I know that \displaystyle t \rightarrow 0 as h approaches to 0. So I have \displaystyle g'(0) = f''(0). But I don't have \displaystyle \frac {f''(0)}{2}, what did I do wrong? 2. Show that g' is continuous on U. First, since \displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} , both f and f' are continuous, and \displaystyle x \neq 0 , so g' must be continuous on \displaystyle U\setminus\{0\} since the sum of two continuous functions is also continuous. (Is this reason enough?) Now, to show that g' is continuous on \displaystyle U \setminus \{ 0 \} . By the Taylor's Theorem, write \displaystyle f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0<s<h Then \displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$$\displaystyle = \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+h^2f''(t)]}{h^2}$
$\displaystyle = \lim _{h \rightarrow 0 } f''(s)-f''(t)- \frac {f(0)}{h^2}$

So as h approaches 0, both s and t approaches 0 as they are sandwiched in between 0 and h, so $\displaystyle f''(s)-f''(t) \rightarrow f''(0)-f''(0) =0$

So all I have left is: $\displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {f(0)}{h^2}$

I must have done something wrong again, because this expression doesn't look converging to g'(0).

Thank you very much for your help!!!
• Dec 7th 2008, 07:49 AM
Laurent
The only problem is a $\displaystyle \frac{1}{2}$ factor in the Taylor's theorem (cf. wikipedia or whereever):

Quote:

My work so far:

1. Use the Taylor's Theorem to show that $\displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$ converges to $\displaystyle \frac{f''(0)}{2}$

Proof.

By the Taylor's Theorem, write $\displaystyle f(x)=f(0)+f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2 \ \ \ \ \ 0<t<x$, since $\displaystyle f(0)=0$, we have $\displaystyle f(x)=f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2$, implies that $\displaystyle f(h)=f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2 \ \ \ \ \ 0<t<h$.

Then $\displaystyle g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf'(0)}{h^2}$
$\displaystyle = \lim _{h \rightarrow 0} \frac {f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2-f'(0)h}{h^2}= \lim _{h \rightarrow 0}{\color{red}\frac{1}{2}}f''(t)$

Now, I know that $\displaystyle t \rightarrow 0$ as h approaches to 0.

So I have $\displaystyle g'(0) = {\color{red}\frac{1}{2}}f''(0)$ because $\displaystyle \color{red}f''$ is assumed to be continuous.

Quote:

2. Show that g' is continuous on U.

First, since
$\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}$, both f and f' are continuous, and $\displaystyle x \neq 0$, so g' must be continuous on $\displaystyle U\setminus\{0\}$ since the sum of two continuous functions is also continuous. (Is this reason enough?)

Now, to show that g' is continuous at 0.

By the Taylor's Theorem, write $\displaystyle f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0<s<h$

Then $\displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$$\displaystyle = \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+{\color{red}\frac{1}{2}}h^2f''(t)]}{h^2}$
$\displaystyle = \lim _{h \rightarrow 0 } f''(s)-{\color{red}\frac{1}{2}}f''(t)- \frac {f(0)}{h^2}$
then, don't forget $\displaystyle f(0)=0$, and you'll be done.