Continuity and periodic functions

[Niles]

Homework Statement

We have a piecewise continuous function and T-periodic function f and we have that:
<br /> F(a) = \int_a^{a + T} {f(x)dx} <br />

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

My attempt so far:

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

<br /> \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}} <br />

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

Now I have that:

<br /> \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}<br />

But where to go from here?

 

[tiny-tim]

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

But where to go from here?

Hi Niles!

Hint: f is T-periodic.

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

 

[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?

 

[tiny-tim]

Then it goes to zero as h -> 0.

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

That was what was puzzling me.

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?

Yup!

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

 

[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

My answers:

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 continuous 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

 

[tiny-tim]

2) I have to show that F is constant

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

Or am I missing something?

 

[Niles]

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

 

[tiny-tim]

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

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

So any constant function is piecewise constant?

 

[Niles]

Hmm, I guess you are right.

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

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

<br /> <br /> \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}<br /> <br />

Is the last step even necessary?

 

[tiny-tim]

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

<br /> <br /> \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}<br /> <br />

Is the last step even necessary?

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.

 

[Niles]

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

<br /> \int_{a + T}^{a + h + T} {f(x) = \int_a^{a + h} {f(x)} } ?<br />

 

[tiny-tim]

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

<br /> \int_{a + T}^{a + h + T} {f(x) = \int_a^{a + h} {f(x)} } ?<br />

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

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

 

[Niles]

Great, thanks!

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