# Thread: Inner Product Involving Integrals

1. ## 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> = $\int^1_0 f(t)g(t)dt$

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.)

2. 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> = $\int^1_0 f(t)g(t)dt$

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: $\langle L(f),g\rangle = \int^1_0 f''(t)g(t)\,dt = \Bigl[f'(t)g(t)\Bigr]_0^1 - \int_0^1f'(t)g'(t)\,dt$ (integration by parts). Now use the fact that g(0) = g(1) = 0.

3. 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> = $\int^1_0 f(t)g(t)dt$

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.)
$=\int_{0}^{1}f''gdx$

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

Then you get

$=\int_{0}^{1}f''gdx=f'g\bigg|_{0}^{1}-\int_{0}^{1}f'gdx$

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

So you end up with

$=\int_{0}^{1}f''gdx=-\int_{0}^{1}f'gdx=$

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

### integrting by parts inner product

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