# Inner Product Involving Integrals

• Mar 6th 2010, 11:56 AM
mathematicalbagpiper
Inner Product Involving Integrals
Let V denote the vector space of all continuous functions :

f:[0, 1] -> R

Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

f(0)=f(1)=0

Now let L: V -> V be defined by

L(f) = f''

Define the inner product <f, g> as follows:

<f, g> = $\displaystyle \int^1_0 f(t)g(t)dt$

Prove that <L(f), g> = <f, L(g)>

(I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)
• Mar 6th 2010, 01:10 PM
Opalg
Quote:

Originally Posted by mathematicalbagpiper
Let V denote the vector space of all continuous functions :

f:[0, 1] -> R

Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

f(0)=f(1)=0

Now let L: V -> V be defined by

L(f) = f''

Define the inner product <f, g> as follows:

<f, g> = $\displaystyle \int^1_0 f(t)g(t)dt$

Prove that <L(f), g> = <f, L(g)>

(I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)

Starting point: $\displaystyle \langle L(f),g\rangle = \int^1_0 f''(t)g(t)\,dt = \Bigl[f'(t)g(t)\Bigr]_0^1 - \int_0^1f'(t)g'(t)\,dt$ (integration by parts). Now use the fact that g(0) = g(1) = 0.
• Mar 6th 2010, 01:14 PM
TheEmptySet
Quote:

Originally Posted by mathematicalbagpiper
Let V denote the vector space of all continuous functions :

f:[0, 1] -> R

Where the 1st and 2nd derivatives exist and are continuous. Suppose further that:

f(0)=f(1)=0

Now let L: V -> V be defined by

L(f) = f''

Define the inner product <f, g> as follows:

<f, g> = $\displaystyle \int^1_0 f(t)g(t)dt$

Prove that <L(f), g> = <f, L(g)>

(I've already proved that L is a linear map. The hint on this part says to use integration by parts twice, but I'm not seeing it. Help? If I can get some guidance on the left-hand side, I can do the right hand side as well since it will be similar. Thanks.)

$\displaystyle <L(f),g>=\int_{0}^{1}f''gdx$

Now as the hint says use integration by parts with
$\displaystyle u=g \implies du=g'dx \text{ and } dv=f''dx \implies v=f'$

Then you get

$\displaystyle <L(f),g>=\int_{0}^{1}f''gdx=f'g\bigg|_{0}^{1}-\int_{0}^{1}f'gdx$

So the middle term is zero becuase $\displaystyle g(1)=g(0)=0$

So you end up with

$\displaystyle <L(f),g>=\int_{0}^{1}f''gdx=-\int_{0}^{1}f'gdx=<f',g>$

Now just integrate by parts one more time to finish the job.