Bilinear map

1. Oct 8, 2009

roam

1. The problem statement, all variables and given/known data
http://img42.imageshack.us/img42/1760/61094288.gif [Broken]

3. 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.

Last edited by a moderator: May 4, 2017
2. Oct 8, 2009

LCKurtz

Your very first step is wrong. It should start:

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

3. Oct 8, 2009

roam

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?

4. Oct 8, 2009

LCKurtz

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

5. Oct 9, 2009

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.

6. Oct 9, 2009

LCKurtz

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

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

7. Oct 9, 2009

LCKurtz

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.

8. Oct 10, 2009

roam

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

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?

9. Oct 10, 2009

LCKurtz

The expression:

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

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.

10. Oct 11, 2009

roam

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 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.

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 $$C[a,b]^*$$.

11. Oct 11, 2009

LCKurtz

So, in terms of

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

what exactly do you need to prove?

12. Oct 12, 2009

roam

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]*?

13. Oct 12, 2009

LCKurtz

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].

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?

14. Oct 12, 2009

roam

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?

15. Oct 12, 2009

LCKurtz

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.

16. Oct 13, 2009

roam

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

Last edited by a moderator: Oct 13, 2009
17. Oct 13, 2009

LCKurtz

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:

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

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

18. Oct 13, 2009

HallsofIvy

Staff Emeritus
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:

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

$$= [f(b)G(b)- \int f'(b)-G(b)dt] - [f(a)G(a)- \int f'(a)-G(a)dt] \neq 0$$

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?

19. Oct 13, 2009

roam

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?

Last edited: Oct 13, 2009
20. Oct 13, 2009

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.