Equality of integrals VS equality of integrands

[Ahmed Mehedi]

Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?

 

[Ahmed Mehedi]

@PeroK @dRic2 @etotheipi

 

[PeroK]

Does $$\int_{t=0}^{\infty}f(t)dt=\int_{t=0}^{\infty}g(t)dt$$ imply $$f(t)=g(t)$$ ?

How could that possibly be true?

 

[etotheipi]

All you need is that the area under the curve from $0$ to $\infty$ is the same. There are a lot of different functions that satisfy this. E.g.

$$
f(x) = \begin{cases}
0, & x< 0 \\
1, & 0\leq x\leq 1 \\
0, & 1< x
\end{cases}$$ and $$
g(x) = \begin{cases}
0, & x< 1 \\
1, & 1\leq x\leq 2 \\
0, & 2< x
\end{cases}$$

 

[Ahmed Mehedi]

@PeroK @dRic2

How could that possibly be true?

Is it possible to prove the equality using calculus of variation ... ? Or may be a special case of it ... ? I am not sure though ...

 

[member 587159]

This is false for many different reasons.

For example @etotheipi gave two different functions.

However, you can also do the following: Consider an integrable function $f$ and change its value in one (or finitely many) points. Let the so obtained be function be $g$. Then clearly $f\neq g$ yet $\int f = \int g$.

For the advanced reader, you are asking if
$$R[0,\infty[ \to \mathbb{R}: f \mapsto \int_{0}^\infty f$$
is injective. This is a linear functional on the space of Riemann-integrable functions on $[0, \infty[$, and a linear functional on an infinite dimensional vector space is never injective.

Note that some partial results do hold:

(1) If $f \geq 0$ and $\int f = 0$ then $f = 0$ almost everywhere.

(2) If $f \geq 0$ and $f$ is continuous with $\int f =0$, then $f=0$ (everywhere).

 

[PeroK]

@PeroK @dRic2

Is it possible to prove the equality using calculus of variation ... ? Or may be a special case of it ... ? I am not sure though ...

Are you really asking that if:
$$\int_{t=0}^{\infty}f(t)dt= 0$$
Then $f(t) = 0$ everywhere?

 

[etotheipi]

There are some specific and limited circumstances under which you can equate certain integrands (I don't know whether it's totally rigorous, so perhaps @PeroK or @Math_QED can advise...). For instance, consider the following statement of Gauss' Law: $$\frac{Q}{\epsilon_0} = \int_V \frac{\rho}{\epsilon_0} \, dV = \oint_S \vec{E} \cdot d\vec{S} = \int_V \nabla \cdot \vec{E} \, dV $$ $$\int_V \frac{\rho}{\epsilon_0} \, dV = \int_V \nabla \cdot \vec{E} \, dV $$ Since this holds for any domain $V$, you may deduce $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$.

I would suspect that if $\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt$ for all possible $a, b$, then you would be able to say $f(t) = g(t)$. But that's quite different to having fixed limits.

 

[PeroK]

I would suspect that if $\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt$ for all possible $a, b$, then you would be able to say $f(t) = g(t)$. But that's quite different to having fixed limits.

If two continuous functions different at a single point, then they differ on an interval. Moreover, if $f(x_0) > g(x_0)$, then $f(x) > g(x)$ on some interval containing $x_0$.

This is needed to extract the Euler-Lagrange equations from the calculus of variations.

 

[member 587159]

I would suspect that if $\int_{a}^{b}f(t)dt=\int_{a}^{b}g(t)dt$ for all possible $a, b$, then you would be able to say $f(t) = g(t)$. But that's quite different to having fixed limits.

This is not quite true (it's true if you ask that $f,g$ are continuous). If they are only Riemann-integrable, best you can get is that they are equal almost everywhere.

Riemann-integral does not see the behaviour at single points, so you can take the example I made earlier in post #6 to get $\int_a^b f = \int_a^b g$ for all $a,b$ yet $f \neq g$.

 

[Ahmed Mehedi]

Are you really asking that if:
$$\int_{t=0}^{\infty}f(t)dt= 0$$
Then $f(t) = 0$ everywhere?

I am very sorry that I can not make my message clear. To be specific my question goes as follows:

Let us consider the following Lagrangian in the context of an optimization problem:

$$L=B\int_{t=0}^{\infty}e^{-\beta t}\frac{c(t)^{1-\theta}}{1-\theta}dt+\lambda \left[ k(0)+\int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}w(t)dt - \int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}c(t)dt \right]$$

After taking the first partial derivative of the above Lagrangian with respect to c(t) (and setting it to zero as first order optimality condition) it is written that

$$Be^{-\beta t} c(t)^{-\theta}-\lambda e^{-R(t)}e^{(n+g)t}=0$$

It seems that they take the first partial derivative of the Lagrangian with respect to c(t) and simply ignore the dt from both sides.

How the second line follows from the first line? They said it can be proved using calculus of variation. But, how can I prove it?

 

[Ahmed Mehedi]

I am very sorry that I can not make my message clear. To be specific my question goes as follows:

Let us consider the following Lagrangian in the context of an optimization problem:

$$L=B\int_{t=0}^{\infty}e^{-\beta t}\frac{c(t)^{1-\theta}}{1-\theta}dt+\lambda \left[ k(0)+\int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}w(t)dt - \int_{t=0}^{\infty} e^{-R(t)}e^{(n+g)t}c(t)dt \right]$$

After taking the first partial derivative of the above Lagrangian with respect to c(t) (and setting it to zero as first order optimality condition) it is written that

$$Be^{-\beta t} c(t)^{-\theta}-\lambda e^{-R(t)}e^{(n+g)t}=0$$

It seems that they take the first partial derivative of the Lagrangian with respect to c(t) and simply ignore the dt from both sides.

How the second line follows from the first line? They said it can be proved using calculus of variation. But, how can I prove it?

@Math_QED @etotheipi

 

[dRic2]

That's a functional derivative, not a partial derivative. It's an other story.

If you want a quick recipe, a functional derivative like $\frac {\delta L} {\delta g}$ where $L =\int h[g]$ can be evaluated by trowing away the integral and calculating $\frac {\partial h[g]} {\partial g}$. Ok don't hate me for this comment.

 

[Ahmed Mehedi]

That's a functional derivative, not a partial derivative. It's an other story.

If you want a quick recipe, a functional derivative like $\frac {\delta L} {\delta g}$ where $L =\int h[g]$ can be evaluated by trowing away the integral and calculating $\frac {\partial h[g]} {\partial g}$. Ok don't hate me for this comment.

May be you are right! Perhaps they were talking about functional derivative and not the partial one. Can you provide me any good quick read regarding functional derivatives?

 

[etotheipi]

LOL this question took a turn...

I don't know if I can help from here on, I've only ever skimmed through the first chapter of a calculus of variations textbook.

 

[Ahmed Mehedi]

LOL this question took a turn...

I don't know if I can help from here on, I've only ever skimmed through the first chapter of a calculus of variations textbook.

HAHA ... I am yet to see the textbook of calculus of variation ... Let alone chapter one ...

 

[dRic2]

Can you provide me any good quick read regarding functional derivatives?

Sorry, I'm not very familiar with functional derivatives. I just worked out some tricks to evaluate them in case of need. All I know comes from pag 54-56 from Lanczos' book on analytical mechanics and from 5 pages of notes by a professor of mine.

 

[Adesh]

Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$
\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions $f$ and $g$ converge to each other as they approach to infinity.

 

[Ahmed Mehedi]

Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$
\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions $f$ and $g$ converge to each other as they approach to infinity.

A good interpretation!

 

[member 587159]

Given
$$\int_{0}^{\infty} f(t) dt = \int_{0}^{\infty} g(t) dt$$ the most we can get is :
$$
\int_{0}^{\infty} f(t) dt - \int_{0}^{\infty} g(t) dt = 0 $$
$$\lim_{x\to \infty} \int_{0}^{x} \left[f(t) -g(t) \right] dt = 0 $$
$$\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$
$$\lim_{x \to \infty} f(x) - g(x) = 0$$
That is to say, functions $f$ and $g$ converge to each other as they approach to infinity.

This is false. Consider $f=0$ and $$g(x) = \begin{cases}1 \quad x \in \mathbb{N} \\ 0 \quad x \notin \mathbb{N}\end{cases}$$
Then $$\int_0^\infty f = 0 = \int_0^\infty g$$ yet $$\lim_{x \to \infty}[ f(x)-g(x)] = -\lim_{x \to \infty} g(x) $$ does not exist.

The flaw happens when you introduce the derivative.

 

[Adesh]

This is false. Consider $f=0$ and $$g(x) = \begin{cases}1 \quad x \in \mathbb{N} \\ 0 \quad x \notin \mathbb{N}\end{cases}$$
Then $$\int_0^\infty f = 0 = \int_0^\infty g$$ yet $$\lim_{x \to \infty}[ f(x)-g(x)] = -\lim_{x \to \infty} g(x) $$ does not exist.

The flaw happens when you introduce the derivative.

I assumed the functions to be continuous.

 

[member 587159]

I assumed the functions to be continuous.

Even then you must justify switching the limits involved, which you didn't. I'm pretty sure the statement is even false for continuous functions.

 

[PeroK]

I assumed the functions to be continuous.

It's still false. There are continuous functions that do not converge to $0$ as $t \rightarrow \infty$ yet the integral exists.

It's a good exercise to find one.

 

[Adesh]

Even then you must justify switching the limits involved, which you didn't. I'm pretty sure the statement is even false for continuous functions.

We can do the switching when they are monotone.

 

[member 587159]

We can do the switching when they are monotone.

If you keep throwing in extra assumptions, eventually you will be right yes...

 

[Adesh]

There are continuous functions that do not converge to 0 as t→∞ yet the integral exists.

I didn’t say they converge to zero, I said they get closer and closer to each other as $x$ goes to $\infty$.

 

[PeroK]

I didn’t say they converge to zero, I said they get closer and closer to each other as $x$ goes to $\infty$.

Which is false. The difference of the functions need not converge to $0$.

Note that if the integral exists and the limit exists, then the limit must be zero. That's true.

 

[Adesh]

Which is false. The difference of the functions need not converge to $0$.

In this case they are coming out be zero, .

 

[Adesh]

@Infrared Did you study from Rudin by yourself? I mean self-study or was your Univeristy good enough from where you can learned it? (Because as far as universities that I know, you cannot learn subjects which require time, like analysis, because teachers are not very interested in teaching).

 

[member 587159]

@Infrared Did you study from Rudin by yourself? I mean self-study or was your Univeristy good enough from where you can learned it? (Because as far as universities that I know, you cannot learn subjects which require time, like analysis, because teachers are not very interested in teaching).

Teachers are only guides. At university, the student is the one that must put in the effort. There are plenty of people that succesfully learned analysis in university, so I'm not sure where your claim comes from.

 

[etotheipi]

AFAIK there are no problems up to here: $$\frac{d}{dx} \lim_{x \to \infty} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$$ but yes I don't think you can generally interchange those limits to get $\lim_{x \to \infty} \frac{d}{dx} \int_{0}^{x} \left[f(t) - g(t) \right] = 0$

 

[Adesh]

so I'm not sure where your claim comes from.

People whom I know (in real life) learned by themselves, teachers only graded them.

 

[member 587159]

People whom I know (in real life) learned by themselves, teachers only graded them.

My opinion is that most of mathematics is self-study. The professor is just there to motivate the concepts and give a first exposure, or to ask questions to.

 

[Adesh]

My opinion is that most of mathematics is self-study. The professor is just there to motivate the concepts and give a first exposure, or to ask questions to.

Then what does it mean when so many people say “Rudin is not good for self-study” ?

 

[member 587159]

Then what does it mean when so many people say “Rudin is not good for self-study” ?

It means that it is a bad book to learn the material from, with which I agree. There are much better self-study books out there.