# Questions about Derivatives and Continuity.

1. Apr 10, 2013

### Bachelier

1. Is this the only example of a function $f(x) \in C^1([0,1])$ with discontinuous derivative

$$f(x) = \begin{cases} x^2 sin(\frac{1}{x}) & \textrm{ if }x ≠ 0 \\ 0 & \textrm{ if }x = 0 \\ \end{cases}$$

It seems this example is over-used. Do we have other examples besides this one in whatever metric space?

2. Also, can a function from a disjoint set be continuous (under the usual metric)?

For instance

$H(x) \ : \ [-∞,-1] \cup [1,∞] → \mathbb{R}$

$\hspace{5 cm} x \mapsto |x|$​

Last edited: Apr 10, 2013
2. Apr 10, 2013

### Bacle2

Take,

f(x)=-x2/2 , x<0

f(0)=0

f(x)=x2/2 , x>0

Then f'(x)=|x| .

Try also using the fact that every a.e. continuous function is Riemann-integrable and

the FThm of Calc.

Last edited: Apr 10, 2013
3. Apr 10, 2013

### Bacle2

Yes; consider a basic open set (a,b) in ℝ . What is its inverse image under f? Check that its

open under the ( I assume you're using) subspace topology of the domain.

EDIT: Like Ivy wrote, the statement disjoint set may be imprecise. I think you mean either a disconnected set, or a set that is not a continuum.

Last edited: Apr 10, 2013
4. Apr 10, 2013

### HallsofIvy

What do you mean by as "disjoint" set? In topology one defines a function, from topological space A to topological space B, to be "continuous" if and only if the inverse image of an open set is open. That is, f:A-> B is continuous if and only if, for any open set X in B, f-1(X) is open in A.

One type of "disjoint" set might be one with the "discrete" topology in which every set is "open". If A has the discrete topology, the f:A-> B for any B is trivially continuous. On the other hand, f:B->A generally will NOT be continuous.

5. Apr 10, 2013

### micromass

I wish to remark that differentiation doesn't necessarily need to make sense in arbitrary metric spaces.

That is continuous.

6. Apr 10, 2013

### Bachelier

everything under the Euclidean metric