1. ## Differentiable FUnctions

a) Prove from definition that if
f, g : R -->R are differentiable functions, then the product function fg is also differentiable.
(b) Prove that if
h : R --->R is a differentiable function that satisties
|h(x)-h(y)|< or = |x-y|^1.5
for all x; y elements of R, then h(x)=0 for all x.
(c) Give an example of a function
k : R --->R such that k is differentiable but not twice differentiable.

2. For functions on the real numbers, the definition of "differentiable" at a point is that the derivative exist at that point. So to prove that "if f and g are differentiable at x= a, then fg is differentiable there", the best thing to do is to follow the derivation of the product rule.
The derivative of fg at x= a is $\displaystyle \displaystyle\lim_{h\to 0}\frac{f(a+h)(g(a+h)- f(a)g(a)}{h}= \lim_{h\to 0}\frac{f(a+h)g(a+h)- f(a)g(a+h)+ f(a)g(a+h)- f(a)g(a)}{h}$ and now separate into two fractions. Use the fact that f and g, separately are differentiable. Also, you will need to use the fact that, since f is differentiable, f is continuous.

For (b), use the mean value theorem: for any x and y, there exist c between x and y, such that h(x)- h(y)= h'(c)(x- y). Take the absolute value of each side and use the fact that $\displaystyle |h(x)- h(y)|\le |x- y|^{1.5}$.

For (c), start with a function that is continuous but not differentiable (there is a well known example) and integrate it!

3. Originally Posted by HallsofIvy
For functions on the real numbers, the definition of "differentiable" at a point is that the derivative exist at that point. So to prove that "if f and g are differentiable at x= a, then fg is differentiable there", the best thing to do is to follow the derivation of the product rule.
The derivative of fg at x= a is $\displaystyle \displaystyle\lim_{h\to 0}\frac{f(a+h)(g(a+h)- f(a)g(a)}{h}= \lim_{h\to 0}\frac{f(a+h)g(a+h)- f(a)g(a+h)+ f(a)g(a+h)- f(a)g(a)}{h}$ and now separate into two fractions. Use the fact that f and g, separately are differentiable. Also, you will need to use the fact that, since f is differentiable, f is continuous.

For (b), use the mean value theorem: for any x and y, there exist c between x and y, such that h(x)- h(y)= h'(c)(x- y). Take the absolute value of each side and use the fact that $\displaystyle |h(x)- h(y)|\le |x- y|^{1.5}$.

For (c), start with a function that is continuous but not differentiable (there is a well known example) and integrate it!
For (b) i obtain $\displaystyle |h'(c)(x- y)|=|h(x)- h(y)|\le |x- y|^{1.5}$.
therefore $\displaystyle |h'(c)(x- y)|\le |x- y|^{1.5}$. Is there any clue how I can get h(x)=0 for all x?

4. Dividing both sides by $\displaystyle |x- y|$, $\displaystyle |h'(c)|\le |x- y|^{.5}$ since that is a positive power, if we take x and y very close together, the right side can be as close as we please to 0. That implies that h'(c) is 0 for all c- that is, that h(x) is a constant. Now show that that constant must be 0.

5. Originally Posted by HallsofIvy
Dividing both sides by $\displaystyle |x- y|$, $\displaystyle |h'(c)|\le |x- y|^{.5}$ since that is a positive power, if we take x and y very close together, the right side can be as close as we please to 0. That implies that h'(c) is 0 for all c- that is, that h(x) is a constant. Now show that that constant must be 0.
So in order to show that the constant must be 0, do we show that h(c) passes through the origin at (0,0)?