Proving Bilinearity of Integrals

[roam]

Homework Statement

http://img42.imageshack.us/img42/1760/61094288.gif

The Attempt at a Solution

Starting with part (a). I need to show that the map is a left-linear form and a right linear form in order to prove that it's bilinear.

For any \alpha, \beta \in \mathbb{R} and any f_i , g_i \in C[a,b]

(i) left-linear form
I(\alpha f_1 + \beta f_2 , g) = (\int_a^{b} \alpha f_1(t) + \int_a^{b} \beta f_2(t)). \int_a^{b} g(t)

= [\alpha \int^b_{a} f_1(t) +\beta \int^b_{a} f_2(t)].\int^b_{a} g(t)

\alpha \int^b_{a} f_1(t)g(t) + \beta \int^b_{a} f_2 (t)g(t) = \alpha I (f_1, g)+\beta I (f_2,g)

(ii) Right-linear form
I(f, \alpha g_1 + \beta g_2) = \int_{a}^{b} f(t).[\alpha \int_{a}^{b} g_1 (t) + \beta \int_{a}^{b} g_2 (t)]

[\int_{a}^{b} f(t) + \alpha \int_{a}^{b} g_1(t)] + [\int_{a}^{b} f(t). \beta \int_{a}^{b} g_2 (t)]

\alpha I (f, g_1) + \beta I (f,g_2)

Is this all I need to show? I'm really not sure if my working here is right. I appreciate it if anyone could correct me if I'm wrong.

 

[LCKurtz]

Your very first step is wrong. It should start:

I(\alpha f_1 + \beta f_2, g)=\int_a^b(\alpha f_1(t) + \beta f_2(t))\cdot g(t)\, dt

 

[roam]

Your very first step is wrong. It should start:

I(\alpha f_1 + \beta f_2, g)=\int_a^b(\alpha f_1(t) + \beta f_2(t))\cdot g(t)\, dt

So, is this correct:

I(\alpha f_1 + \beta f_2, g)=\int_a^b(\alpha f_1(t) + \beta f_2(t))\cdot g(t)\, dt

= \alpha \int_a^b (f_1 (t) g(t)) + \beta \int_a^b (f_2 (t) , g(t))

= \alpha I (f_1, g)+\beta I (f_2,g)

Is this the right representation?

 

[LCKurtz]

Your second lined is bollixed. Spare parentheses, extraneous commas, and no dt's.

 

[roam]

oops

I(\alpha f_1 + \beta f_2, g)=\int_a^b(\alpha f_1(t) + \beta f_2(t))\cdot g(t)\, dt

= (\alpha \int_a^b (f_1 (t) , g(t))dt) + (\beta \int_a^b (f_2 (t) , g(t)) dt)

= \alpha I (f_1, g)+\beta I (f_2,g)

Is it better now?

If so, could you show me how to do part (b)? I looked up my textbook and a few website but couldn't figure out how to show using the right definitions that the bilinear form is non-degenerate. I appreciate any guidance.

 

[LCKurtz]

No, it isn't better. What would you make of an integral like this:

\int_a^b \sin(x),\cos(x)\ dx

 

[LCKurtz]

oops


If so, could you show me how to do part (b)? I looked up my textbook and a few website but couldn't figure out how to show using the right definitions that the bilinear form is non-degenerate. I appreciate any guidance.

The first step for part b would be to write the definition for what it means for this bilinear form to be non-degenerate. Then at least you know exactly what you have to prove or disprove.

 

[roam]

No, it isn't better. What would you make of an integral like this:

\int_a^b \sin(x),\cos(x)\ dx

\int_a^b \sin(x),\cos(x)\ dx = -cos(b)+cos(a), sin(b)-sin(a)

But I don't know how exactly it can be done with

(\alpha \int_a^b (f_1 (t) , g(t))dt) + (\beta \int_a^b (f_2 (t) , g(t)) dt)

Okay, for part (b), I don't know what definition to use. I looked up some websites and they talk about "isomorphism", but we still haven't come across this term yet. Is there anything simpler to be do here?

 

[LCKurtz]

The expression:

<br /> \int_a^b \sin(x),\cos(x)\ dx<br />

makes no sense. The "," is not an arithmetic operator, and commas make no sense inside of integrals. So your middle step and proposed meaning are still nonsense.

For part b you need to do two things to get started:

1. Finish the statement of this definition: A bilinear form B(x,y) over a vector space V is non-degenerate if ... Presumably that is in your course textbook.

2. State what that means for your bilinear form I(f,g) in terms of its definition.

You have no hope of proving or disproving the statement before you have a clear understanding of what the statement is.

 

[roam]

The expression:

<br /> \int_a^b \sin(x),\cos(x)\ dx<br />

makes no sense. The "," is not an arithmetic operator, and commas make no sense inside of integrals. So your middle step and proposed meaning are still nonsense.

OK, fixed!

(\alpha \int_a^b f_1 (t) . g(t)dt) + (\beta \int_a^b f_2 (t) . g(t) dt)

= \alpha I (f_1, g)+\beta I (f_2,g)

For part b you need to do two things to get started:

1. Finish the statement of this definition: A bilinear form B(x,y) over a vector space V is non-degenerate if ... Presumably that is in your course textbook.

...for each x \in V \ \{ 0_V \}, the linear maps V \rightarrow F given by x \mapsto B(x,y) and x \mapsto B(y,x) are non-zero in V*.

Here by V* I mean the dual space of V, set of all linear maps from V to F.

This definition isn't from my textbook. Textbook's definition was to do with rank and dimension, which is not the one you are refferring to.

2. State what that means for your bilinear form I(f,g) in terms of its definition.

Bilinear I(f,g) is non-degenerate if for each x \in C[a,b]\{0_{C[a,b]} \} the linear maps given by x \mapsto I(f,g) and x \mapsto I(g,f) are non-zero in C[a,b]^*.

 

[LCKurtz]

Bilinear I(f,g) is non-degenerate if for each x \in C[a,b]\{0_{C[a,b]} \} the linear maps given by x \mapsto I(f,g) and x \mapsto I(g,f) are non-zero in C[a,b]^*.

So, in terms of

I(f,g) = \int_a^b f(t)g(t)\, dt

what exactly do you need to prove?

 

[roam]

So, in terms of

I(f,g) = \int_a^b f(t)g(t)\, dt

what exactly do you need to prove?

I don't know since I have no worked examples, but I guess I must show that the linear maps

t \mapsto \int_a^b f(t)g(t) dt and the t \mapsto \int_a^b f(t)g(t) dt

are non-zero in C[a,b]*?

 

[LCKurtz]

I don't know since I have no worked examples, but I guess I must show that the linear maps

t \mapsto \int_a^b f(t)g(t) dt and the t \mapsto \int_a^b f(t)g(t) dt

are non-zero in C[a,b]*?

Those aren't maps in C^*. They don't even make any sense because there is no t in the right side; it is a dummy variable. And your domain must be C[a,b].

It might help you to phrase the definition differently. Try this:

A bilinear form B over a vector space V is said to be non-degenerate when

1. if B(x,y)=0 for all x in V then y=0 and
2. if B(x,y)=0 for all y in V then x=0

Can you interpret that in your setting for I(f,g)? So what do you have to prove?

 

[roam]

Those aren't maps in C^*. They don't even make any sense because there is no t in the right side; it is a dummy variable. And your domain must be C[a,b].

It might help you to phrase the definition differently. Try this:

A bilinear form B over a vector space V is said to be non-degenerate when

1. if B(x,y)=0 for all x in V then y=0 and
2. if B(x,y)=0 for all y in V then x=0

Can you interpret that in your setting for I(f,g)? So what do you have to prove?

Is this it:

The bilinear form I on a vector space C[a,b] is non-degenerate

(i) If I(f,g)=0 \forall f \in C[a,b] then g=0

(ii) And if I(f,g)=0 \forall g \in C[a,b] then f=0

How are we exactly meant to show that if
I(f,g) = \int_a^b f(t)g(t)\, dt = 0
for any f, then g=0? Any hints?

 

[LCKurtz]

Is this it:

The bilinear form I on a vector space C[a,b] is non-degenerate

(i) If I(f,g)=0 \forall f \in C[a,b] then g=0

(ii) And if I(f,g)=0 \forall g \in C[a,b] then f=0

How are we exactly meant to show that if
I(f,g) = \int_a^b f(t)g(t)\, dt = 0
for any f, then g=0? Any hints?

Yes, that is what you need to show. My hint would be to try an indirect argument. Suppose that g(t₀) isn't zero for some t₀ and see if you can figure out some f such that

I(f,g) = \int_a^b f(t)g(t)\, dt \ne 0

thus giving a contradiction. Don't forget to justify any arguments by stating explicitly any results you use from calculus.

 

[roam]

Yes, that is what you need to show. My hint would be to try an indirect argument. Suppose that g(t₀) isn't zero for some t₀ and see if you can figure out some f such that

I(f,g) = \int_a^b f(t)g(t)\, dt \ne 0

thus giving a contradiction. Don't forget to justify any arguments by stating explicitly any results you use from calculus.

I don't understand what is meant by "results you use from calculus". Which results are you referring to?

 

[LCKurtz]

I don't understand what is meant by "results you use from calculus". Which results are you referring to?

I was just repeating what was stated in your problem. You aren't going to be able to use integration by parts because you don't have formulas for the functions. You need to think more theoretically about general properties of continuous functions and integrals. Properties that you would have learned in calculus.

All you are assuming is that g(t₀) isn't zero for some t₀ in [a,b]. Given that information about g, you need to show how you can construct a function f(t) such that:

I(f,g) = \int_a^b f(t)g(t)\, dt \ne 0<br />

It isn't that difficult once you start thinking about it.

 

[HallsofIvy]

I don't understand what is meant by "results you use from calculus". Which results are you referring to?

I accidently editted roam's last response when I meant to copy it. I think I have restored it correctly.


The only calculus results I can think of is that we use integration by parts for integrals of this form, which are a product of two functions:

I(f,g) = \int_a^b f(t)g(t)\, dt \neq 0

= [f(b)G(b)- \int f&#039;(b)-G(b)dt] - [f(a)G(a)- \int f&#039;(a)-G(a)dt] \neq 0

By G(t) I mean any antiderivative of g(t) and f' the derivative of f.

How does this exactly help us with the contradiction?

 

[roam]

I was just repeating what was stated in your problem. You aren't going to be able to use integration by parts because you don't have formulas for the functions. You need to think more theoretically about general properties of continuous functions and integrals. Properties that you would have learned in calculus.

All you are assuming is that g(t₀) isn't zero for some t₀ in [a,b]. Given that information about g, you need to show how you can construct a function f(t) such that:

<br /> I(f,g) = \int_a^b f(t)g(t)\, dt \ne 0<br />

It isn't that difficult once you start thinking about it.

Yes I'm thinking more theoretically but I can't think of any other useful theorems/properties from integral calculus. The only other I can think of is the "mean value theorem for integrals":

\exists t^* \in [a,b] such that

I(f,g) = \int_a^b f(t)g(t)\, dt

= [f(t^*)g(t^*)](b-a)

Now if we could assume that t^* has the same value as t₀, and f(t₀) doesn't equal to zero, since g(t₀) isn't 0 we could end up with:

\int_a^b f(t)g(t)\, dt = [f(t^*)g(t^*)](b-a) \neq 0

But I know that this theorem doesn't help us either. What other properties can we consider?

 

[LCKurtz]

Instead of trying to think of theorems from calculus you can use, why don't you just focus on trying to see how to construct an f(t) that makes the integral nonzero. If you need any calculus theorems, that will be apparent soon enough.

 

[roam]

Instead of trying to think of theorems from calculus you can use, why don't you just focus on trying to see how to construct an f(t) that makes the integral nonzero. If you need any calculus theorems, that will be apparent soon enough.

How can we construct such an f(t)? I absolutely have no idea! The question does not even define the functions f(t) and g(t)... I'm very confused

 

[LCKurtz]

How can we construct such an f(t)? I absolutely have no idea! The question does not even define the functions f(t) and g(t)... I'm very confused

Integrals represent areas under graphs. You know g(t₀) is nonzero, say it's positive. Is the area under g(t) between a and b positive? Probably not; you don't know enough about g. What could go wrong? But can you multiply it by some continuous f of your choice to make the area positive? Draw some pictures.