Homework Help: Continuity and periodic functions

1. Sep 8, 2008

Niles

1. The problem statement, all variables and given/known data
We have a piecewise continous function and T-periodic function f and we have that:
$$F(a) = \int_a^{a + T} {f(x)dx}$$

I have to show that F is diferentiable at a if f is continous at a.

My attempt so far:

I have showed that F is continous for all a. If we look at one piece where f is continous for a, we have that if F'(a) = f(a):

$$\frac{1}{h}\int_a^{a + h} {f(x)dx = \frac{{\int_x^{a + h} {f(x)dx - } \int_x^a {f(x)dx} }}{h}}$$

When h -> 0, then the above goes to f(a) - this is just the Fundemental Theorem of Calculus.

Now I have that:

$$\frac{{F(a + h) - F(a)}}{h} = \frac{{\int_{a + h}^{a + h + T} {f(x)dx - } \int_a^{a + T} {f(x)dx} }}{h} = \frac{{\int_{a + T}^{a + h + T} {f(x)dx - } \int_a^{a + h} {f(x)dx} }}{h}$$

But where to go from here?

2. Sep 8, 2008

tiny-tim

Hi Niles!

Hint: f is T-periodic.

(hmm … doesn't this work even if f is not continuous? )

3. Sep 8, 2008

Niles

Then it goes to zero as h -> 0. The limit exists, and hence F is differnetiable at all points a, where f is diff. - correct?

Also, does this mean that F'(a) = 0?

4. Sep 8, 2008

tiny-tim

Actually, it's 0 anyway, isn't it?

That was what was puzzling me.
Yup!

though the actual value was obvious from the definition, wasn't it?

5. Sep 8, 2008

Niles

Great - I have two final (easy!) questions.

1) I have to conclude that F is piece-wise constant.

2) I have to show that F is constant

1) Ok, it's kinda obvious that F must be constant. But I can't find the argument for F being piecewise constant, because I have shown that F is continous for all a in a previous question. Just because f is P.C., then it doesn't mean that F is it also.

2) If F is constant (and periodic and piecewise continous), then it must be constant for all a, since it is bounded. This question was easy - #1 is annoying me :grumpy:

6. Sep 8, 2008

tiny-tim

Isn't that obvious from the previous formula for F(a + h) - F(a)?

Or am I missing something?

7. Sep 8, 2008

Niles

Question nr. 2 was the easy one. Question nr. 1 was what I couldn't do

8. Sep 8, 2008

tiny-tim

Doesn't piecewise constant just mean a multi-step step-function?

So any constant function is piecewise constant?

9. Sep 8, 2008

Niles

Hmm, I guess you are right.

I have one final question, then I'll let you off the hook (I made a joke :rofl:).

In my first post, in the last expression, I changed the limits of integration - here it is:

$$\frac{{F(a + h) - F(a)}}{h} = \frac{{\int_{a + h}^{a + h + T} {f(x)dx - } \int_a^{a + T} {f(x)dx} }}{h} = \frac{{\int_{a + T}^{a + h + T} {f(x)dx - } \int_a^{a + h} {f(x)dx} }}{h}$$

Is the last step even necessary?

10. Sep 8, 2008

tiny-tim

Hi Niles!

Yes, I think it is …

it's only the periodicity of f that makes it zero …

and the periodicity doesn't obviously apply until the last step.

11. Sep 8, 2008

Niles

Hmm, I can't quite see that. Do you mean that:

$$\int_{a + T}^{a + h + T} {f(x) = \int_a^{a + h} {f(x)} } ?$$

Last edited: Sep 8, 2008
12. Sep 8, 2008

tiny-tim

Yes, because the substitution y = x + T makes it

$$\int_a^{a + h} f(y)\,dy\ =\ \int_a^{a + h} f(x)\,dx$$

13. Sep 9, 2008

Niles

Great, thanks!

http://www.thefunones.com/costume_rental_shark.jpg [Broken]

Last edited by a moderator: May 3, 2017