Show that the integral tends to zero

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Homework Statement



How do I show that this integral tends to zero as b tends to a ?

Homework Equations



[itex]\int_a^b \int_0^a (f(a)-f(y))(x-y)dy(b-x)dx[/itex]

The Attempt at a Solution



I'm not sure how to start.
Any hints or ideas will be very much appreciated.
We are given that function f is continuous over the domain.

Thank you.
 

Answers and Replies

  • #2
jbunniii
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The inner integral ##\int_{0}^{a} (f(a) - f(y)) (x-y) dy## is a function of ##x##. Can you show that it is bounded? In other words, there is some ##M## such that for all ##x \in [a,b]##,
$$\left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| \leq M$$
If so, then you will be able to argue as follows:
$$\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
M \int_{a}^{b} |b-x| dx$$
and the result should follow easily.
 
  • #3
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Thank you very much.

To show that the inner integral is bounded, I first simplified
[itex]|\int_0^a (f(a)-f(b))(x-y)dy |[/itex] since f(a) is independent of x.

So I split the integral into 2 parts:
[itex]\int_0^a f(a)(x-y)dy-\int_0^a f(b)(x-y)dy[/itex]

Evaluating the first part:
[itex]\int_0^a f(a)(x-y)dy=\frac{1}{2}f(a)(2xa+a^2)[/itex]

So I have to show that there is an [itex]M[/itex] such that for [itex]x\in (a,b)[/itex]
[itex]|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(b)(x-y)dy|\le M[/itex]

Would M be a constant or function of x? I'm a bit confused.
 
  • #4
jbunniii
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Would M be a constant or function of x? I'm a bit confused.
It needs to be independent of ##x##. So far your work looks good. You can now apply the triangle inequality to get
$$\left|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(b)(x-y)dy\right| \leq
\left|\frac{1}{2}f(a)(2xa+a^2)\right| + \left|\int_0^a f(b)(x-y)dy\right|$$
The first term should be easy to bound since you know ##a \leq x \leq b##. The second term should be similarly easy if you note that ##f(b)## is a constant and that ##|\int \ldots| \leq \int |\ldots|##.
 
  • #5
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Thank you.

I see, so I have:
[itex]|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(y)(x-y)dy|\le M[/itex]
Using the triangle inequality, we have:
[itex]|\frac{1}{2}f(a)(2xa+a^2)|+|\int_0^a f(y)(x-y)dy|\le M[/itex]
The first term is bounded by
[itex]\frac{1}{2}f(a)(2ba+a^2)[/itex]
So, we can write:
[itex]|\frac{1}{2}f(a)(2ab+a^2)|+\int_0^a |f(y)(x-y)|dy\le M[/itex]

Now, for the second term, I'm still unsure, if we can't integrate it, is it still easy to see what it is bounded by?

I just noticed that in my second post, all f(b) should actually read f(y).
 
  • #6
jbunniii
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Thank you.

I see, so I have:
[itex]|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(y)(x-y)dy|\le M[/itex]
Using the triangle inequality, we have:
[itex]|\frac{1}{2}f(a)(2xa+a^2)|+|\int_0^a f(y)(x-y)dy|\le M[/itex]
The first term is bounded by
[itex]\frac{1}{2}f(a)(2ba+a^2)[/itex]
So, we can write:
[itex]|\frac{1}{2}f(a)(2ab+a^2)|+\int_0^a |f(y)(x-y)|dy\le M[/itex]
Well, you shouldn't write ##\leq M## yet because you haven't shown that. Instead, you can write
$$\begin{align}
\left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
&\leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right|\\
\end{align}$$
Now, you said the first term is bounded by ##\frac{1}{2}f(a)(2ba + a^2)##, but even after adding the missing absolute value signs, that isn't necessarily true if ##a## and ##b## are allowed to be negative. Consider for example ##a=-1##, ##b=0##. However, you can get a bound by applying the triangle inequality again:
$$\begin{align}
\left| \frac{1}{2}f(a)(2xa + a^2)\right|
&\leq \frac{1}{2}|f(a)|(|2xa| + |a^2|)\\
&= \frac{1}{2}|f(a)|(2|a|\cdot|x| + |a|^2)\\
\end{align}$$
Now find a bound for ##|x|## and that takes care of the first term.

For the second term, you need to use the fact that ##f## is continuous, hence bounded on ##[0,a]## (why?), i.e. there exists some bound ##B## such that ##|f(y)| \leq B## for all ##y \in [0,a]##. Now, can you find a bound for ##|x-y|##? Remember that ##y \in [0,a]## and ##x \in [a,b]##.
 
  • #7
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We have: [itex]|x|\le max(|a|,|b|)[/itex].

For the second term:
We can say that f is continuous and bounded on [0,a] since we can take

[itex]B=max(|f(a)|,f(0)|[/itex]
so that
[itex]|f(y)|\le B\,\,\,\, \forall\,\, y\in[0,a][/itex]

I think |x-y| must be bounded by b. is that right?
It's the biggest difference it can take for [itex]x\in[a,b]; y\in[0,a][/itex].

So does that mean that the initial double integral is bounded by M*b ?
 
Last edited:
  • #8
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The inner integral ##\int_{0}^{a} (f(a) - f(y)) (x-y) dy## is a function of ##x##. Can you show that it is bounded? In other words, there is some ##M## such that for all ##x \in [a,b]##,
$$\left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| \leq M$$
If so, then you will be able to argue as follows:
$$\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
M \int_{a}^{b} |b-x| dx$$
and the result should follow easily.

I think you guys are making a mistake you fill in f(b) while it should be f(y) in the remainder of your calculations.
 
  • #9
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Thank you, yes you're right.
The f(b) should be f(y) in post numbers 3 and 4.
We have fixed this in the following posts.
 
  • #10
jbunniii
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For the second term:
We can say that f is continuous and bounded on [0,a] since we can take

[itex]B=max(|f(a)|,f(0)|[/itex]
No, the max won't necessarily occur at an endpoint of the interval. You have to invoke the theorem that a function which is continuous on a closed bounded interval achieves a maximum and a minimum on the interval, hence it is bounded. The theorem doesn't say where the max or min occurs, but it does guarantee that there is some ##C## (to use a different letter from the other bounds we already found) such that ##|f(x)| \leq C## for all ##x \in [0,a]##.
I think |x-y| must be bounded by b. is that right?
Not necessarily. Is ##b## even guaranteed to be positive? It's not stated in the problem anywhere. However, you can use the good old triangle inequality again to get a bound:
##|x-y| \leq |x| + |y| \leq ???##
 
  • #11
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Oh, I see now why f is bounded. thank you!

I've just checked and we have that [itex]a[/itex] and [itex]b[/itex] are positive.
I should have stated that in the problem.
Also, [itex]b \ge a[/itex].

So, can I say that
[itex]|x-y|\le |x|+|y| \le a+b[/itex] ?
 
  • #12
jbunniii
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Oh, I see now why f is bounded. thank you!

I've just checked and we have that [itex]a[/itex] and [itex]b[/itex] are positive.
I should have stated that in the problem.
Also, [itex]b \ge a[/itex].

So, can I say that
[itex]|x-y|\le |x|+|y| \le a+b[/itex] ?
Yes, that works fine. Now what happens when you apply all the bounds that you've found to the original inequality?
 
  • #13
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[itex]
\left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
\leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right| [/itex]

[itex]\le \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b)[/itex]
(with a>0)

What then happens with the original inequality with the double integral?
[itex]\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx[/itex] ?
 
  • #14
jbunniii
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[itex]
\left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
\leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right| [/itex]

[itex]\le \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b)[/itex]
(with a>0)

What then happens with the original inequality with the double integral?
[itex]\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx[/itex] ?
Let's give your bound a name: ##M = \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b)##. Then you have shown that
$$\left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| \leq \left|\frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a} f(y)(x-y)dy \right| \leq M$$
So apply that to your original integral:
$$\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
\int_{a}^{b} \left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| |b-x| dx \leq ???$$
 
  • #15
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We need to find a bound for
[itex]|b-x|[/itex] for [itex]x[/itex] between [itex]a[/itex] and [itex]b[/itex]

So,
[itex]\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
\int_{a}^{b} \left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| |b-x| dx \leq M(b-a)[/itex]
Is that right?

(Thank you by the way, very much appreciated.)
 
  • #16
jbunniii
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Yes, that looks good. Now what can you conclude as ##b \rightarrow a##?
 
  • #17
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As [itex]b \rightarrow a[/itex], we have that the upper bound tends to zero... ?
 
  • #18
jbunniii
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As [itex]b \rightarrow a[/itex], we have that the upper bound tends to zero... ?
Right, so what does that imply about ##\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right|## as ##b \rightarrow a##?
 
  • #19
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This implies that it tends to zero...?
 
  • #20
jbunniii
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This implies that it tends to zero...?
Are you asking me, or telling me?

You have an inequality of this type:
$$0 \leq |\text{ function of }a\text{ and }b| \leq M(b-a)$$
If ##b \rightarrow a##, then the right hand side tends toward zero. The left side is already zero. So the guy in the middle is squeezed between zero and something which approaches zero. What can you conclude?
 
  • #21
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I was asking just to make sure :)

It implies that the guy in the middle must also approach 0.
 
  • #22
jbunniii
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OK, looks like you got it.
 

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