# Piecewise smooth and piecewise continuous

1. Aug 24, 2008

### Niles

1. The problem statement, all variables and given/known data
When a function is piecewise smooth, then f and f' (the derivative of f) are piecewise continuous.

In my book they mention "a function f, which is continuous and piecewise smooth". How can f be both continuous and piecewise continuous?

2. Aug 24, 2008

### CompuChip

If it is continuous, it is piecewise continuous (in one big piece).
If it is piecewise smooth, then it needn't be piecewise continuous.

3. Aug 24, 2008

### HallsofIvy

Staff Emeritus
For example, f(x)= |x| is "continuous and piecewise differentiable": it is continuous for all x and differentiable every where except at x= 0 so differentiable on the "pieces" $(-\infty, 0)$ and $(0, \infty)$.

4. Aug 24, 2008

### Niles

Great, thanks to both of you.

5. Aug 29, 2008

### Niles

Hmm wait, I'm still having some difficulties with this definition of a function being piecewise smooth. I understand your examples - but still I can't see how a function can be both piecewise continuous and continuous at the same time.

If f is continuous, then f is piecewise continuous in one big piece, then there really isn't a difference between the two terms? I mean, in your example HallsOfIvy, f(x) = |x| is not piecewise continuous, but continuous. So if I use the definition from my book, then it cannot be piecewise smooth? (Although I can see why it is piecewise smooth, but I don't like the definition in my book).

Last edited: Aug 29, 2008
6. Aug 29, 2008

### HallsofIvy

Staff Emeritus
perhaps what is confusing you is that in this definition they are thinking of "piecewise continuous" as including "continuous". That is, every continuous function is "piecewise" continuous.

7. Aug 29, 2008

### Niles

From Wolfram MathWorld (http://mathworld.wolfram.com/PiecewiseContinuous.html): [Broken] "A function or curve is piecewise continuous if it is continuous on all but a finite number of points at which certain matching conditions are sometimes required."

Is it proper to say that in this case, this "finite number" is just zero?

Last edited by a moderator: May 3, 2017
8. Aug 30, 2008

### CompuChip

Exactly.

Another way to look at it:
The intuitive idea of that definition is, that there exist points $a = x_0, x_1, x_2, \cdots, x_n = b$ and functions $f_1, f_2, \cdots, f_n$ such that:
• $f_i$ is defined on ${[x_{i-1}, x_i[}$
• All $f_i$ are continuous
• The original function f on [a, b[ is given by: $$f(x) = f_i(x)$$ if $x \in {[x_{i-1}, x_i[}$.
For example, f(x) = |x| can also be written as
$$f(x) = \begin{cases} f_1(x) \equiv -x & \text{ if } x < 0 \\ f_2(x) \equiv x & \text{ if } x \ge 0 \end{cases}$$.
The $f_i$ are the "pieces" that make up the function. If the function is continuous, then one continuous piece is enough.

9. Aug 30, 2008

### Niles

Great, I got it now! Thanks to both of you.

Have a nice weekend.

10. Aug 30, 2008

### Niles

Ok, I have a new question similar to this one, so I thought it would make more sense just posting it in here.

Does the same discussion go for piecewise smooth functions? I mean, as we speak, I am writing the Fourier-series for f(x) = |sin(x)|, and then I thought: "This is a periodic function, but not piecewise smooth? Then how can the function have a Fourier-series?"

But it this the same as we talked about earlier? That it is piecewise smooth in "one big piece"?

11. Aug 30, 2008

### CompuChip

It is piecewise smooth. It is given by:
f(x) = -sin(x) on [-pi, 0) and sin(x) on (0, pi];
both pieces are smooth (e.g. it is not differentiable in finitely many points only).

12. Aug 30, 2008

### Niles

Why is it not defined in x=0?

13. Aug 31, 2008

### HallsofIvy

Staff Emeritus
It is. That was a typo. f(x) is -sin(x) on [-pi, 0) and sin(x) on [0, pi].

14. Aug 31, 2008

### Niles

But on the interval [-pi;pi], |sin(x)| is both continous and C1. A x=0, the limit from right and left are the same, so no discontinuity here. And the same with the derivative of sine.

So is this "finite number" again zero?

15. Aug 31, 2008

### d_leet

The derivative is not continuous at 0, from the right it approaches 1, from the left it approaches -1.

16. Sep 1, 2008

### CompuChip

So, |sin(x)| is continuous (therefore, piecewise continuous) but it is not differentiable (though it is piecewise smooth*) on, say, [-pi, pi].

* smooth meaning: infinitely often differentiable.

17. Sep 1, 2008

### Niles

Ahh yes, of course. But you mean that from the right the limit is -1 and from the left it approaches 1, right?

Ok, I get it now. Thanks to everybody!

18. Sep 1, 2008

### CompuChip

If you come from the right, you are moving along the graph of sin'(x) = cos(x) which goes to +1 as you approach zero.
If you come from the left, you are moving along the graph of -sin'(x) = -cos(x) which approaches -1.

19. Sep 1, 2008

### Niles

Of course, I was integrating.. wow, I really need to get some more sleep and pay more attention, so this wont happen again. Thanks for being patient.