1. ## [SOLVED] bilinear mapping

Assume a<b. $C[a,b]$ denotes the set of all continuous functions $[a,b] \rightarrow \mathbb{R}$ and that all continuous functions $[a,b] \rightarrow \mathbb{R}$ are integrable. So we have a function $I: C[a,b] \times C[a,b] \rightarrow \mathbb{R}$ given by $I(f,g)= \int^b_a f(t)g(t)dt$.

(I have already proved and concluded that I is a bilinear form).

Here's the question:

Is this bilinear form non-degenerate? Justify your answer by stating explicitly any results you use from calculus.

My Attempt

The bilinear form I is non-degenerate

1. If I(f,g)=0 $\forall f \in C[a,b]$ then g=0
2. If I(f,g)=0 $\forall g \in C[a,b]$then f=0

I personally think this bilinear form is non-degenerate. I will try to give a proof by contradiction. For a contradiction suppose $I(f,g) = \int_a^b f(t)g(t)\, dt = 0$ but $g \neq 0$.

I'm not sure how to carry on and finish this proof from this point. I don't know how to use calculus here... I know that integrals represent areas under the graphs. And in this case we know know $g(t) \neq 0$.
And f is not defined, if we define f = sin(t) and a=-1, b=1 we get

$
\int^1_{-1} sin(t) =0
$

But this is a counter example to what we are trying to prove!

I have to submit this tomorrow, so I appreciate it if anyone could please help me complete this proof...urgently

2. Originally Posted by Roam
Assume a<b. $C[a,b]$ denotes the set of all continuous functions $[a,b] \rightarrow \mathbb{R}$ and that all continuous functions $[a,b] \rightarrow \mathbb{R}$ are integrable. So we have a function $I: C[a,b] \times C[a,b] \rightarrow \mathbb{R}$ given by $I(f,g)= \int^b_a f(t)g(t)dt$.

(I have already proved and concluded that I is a bilinear form).

Here's the question:

Is this bilinear form non-degenerate? Justify your answer by stating explicitly any results you use from calculus.

My Attempt

The bilinear form I is non-degenerate

1. If I(f,g)=0 $\forall f \in C[a,b]$ then g=0
2. If I(f,g)=0 $\forall g \in C[a,b]$then f=0

I personally think this bilinear form is non-degenerate. I will try to give a proof by contradiction. For a contradiction suppose $I(f,g) = \int_a^b f(t)g(t)\, dt = 0$ but $g \neq 0$.

I'm not sure how to carry on and finish this proof from this point. I don't know how to use calculus here... I know that integrals represent areas under the graphs. And in this case we know know $g(t) \neq 0$.
And f is not defined, if we define f = sin(t) and a=-1, b=1 we get

$
\int^1_{-1} sin(t) =0
$

But this is a counter example to what we are trying to prove!

I have to submit this tomorrow, so I appreciate it if anyone could please help me complete this proof...urgently
You may use the following easy

Lemma: if f is a continuous function in (a,b) and f(w) > 0 for some w there then there exists a neighborhood J= (x-e, x+e) in (a,b), with e > 0, s.t. f(x) > 0 fo all x in J

Now, supose I(f,g) = 0 for all f in C[0,1] ==> in particular, I(g,g) = 0 ==>
INT{0,1}[g^(x) dx] = 0 ; if there's some point w in (0,1) s.t. g(w) is not zero then g^2(w) > 0 in some neighborhood J = (c,d) of w ==> 0 < INT{c,d}[g^2(x) dx] <= INT{0,1}[g^2(x)dx] since we know that the Riemann integral of a positive function in an interval of positive length is positive.

Tonio

3. Is this bilinear form even non-degenerate?

4. Originally Posted by Roam
Is this bilinear form even non-degenerate?

Read again carefully my answer: I think the solution is clear there.

Tonio

5. Originally Posted by Roam

1. If I(f,g)=0 $\forall f \in C[a,b]$ then g=0
This is known as the fundamental lemma of calculus of variations.
Fundamental lemma of calculus of variations - Wikipedia, the free encyclopedia

6. Originally Posted by tonio
Read again carefully my answer: I think the solution is clear there.

Tonio
So it IS non-degenerate since the integral is equal to zero?