1. ## Porperty of B-spline

Hello,

Assuming that a B-spline $N_{m}$ is defined for some $m \in \mathbb{N}$ then the B-spline $N_{m+1}$ is defined by:

$N_{m+1}(x)=\int_{0}^{1}N_{m}(x-t))dt$

How can I, through induction, show that:

$\int_{-\infty}^{\infty}N_{m}(x)dx=1$

Suggestions would be greatly appreciated.

2. How is $N_1(x)$ defined?

3. Hi

Working on the same problem. So far I have shown that the property is true for m = 1. My line of thought is then to show that the property is true for m+1. I am however struggling with a double integral.

I do not remember exactly how N_m is characterized. That makes it a bit hard to integrate. (I know that there is an explicit formula for the B-splines, but good luck integrating that.) I was thinking that there must be a better way to do the integration.

4. Hi HallsofIvy

N_1(x) is the characteristic function from [0,1].

5. ## B-spline

Hello,

So we are considering the statement:

$S_{m}: \hspace{0,5cm}\int_{-\infty}^{\infty}N_{m}(x)dx=1$

As ase mentioned the statement $S_{1}$ is true for $m=1$. This establishes the basis for the induction.

Now we assume that $S_{m}$ is true for a fixed value of $m$. Using the induction hypothesis we want to derive $S_{m+1}$. We get:

$\int_{-\infty}^{\infty}N_{m+1}(x)dx=\int_{-\infty}^{\infty}\int_{0}^{1}N_{m}(x-t)dt$

From here I'm not sure. This is where Ase got stuck (I presume). Suggestions are appreciated greatly. Thanks.

6. Originally Posted by surjective
Hello,

Assuming that a B-spline $N_{m}$ is defined for some $m \in \mathbb{N}$ then the B-spline $N_{m+1}$ is defined by:

$N_{m+1}(x)=\int_{0}^{1}N_{m}(x-t))dt$

How can I, through induction, show that:

$\int_{-\infty}^{\infty}N_{m}(x)dx=1$

Suggestions would be greatly appreciated.
Are you familiar with the convolution of functions? because if so this problem is easy since $N_m = f\ast N_{m-1}$ where $f=1_{[0,1]}$ and it's well known that since both $f$ and $N_{m-1}$ are in $L^1(\mathbb{R} )$ then $\int_{\mathbb{R} } N_m = \left( \int_{\mathbb{R} } f \right) \left( \int_{\mathbb{R} } N_{m-1} \right) = 1$

7. ## Property revisited

Thanks for the reply. Not sure if I get the convultion-idea. I know the definition of convultion and that by definition:

$N_{m}=N_{1}*N_{1}*N_{1}*\cdots *N_{1}$

Could you maybe outline your comment one more time.

Furthermore, I was wondering if the intended could be shown via induction? Is that possible?

8. Originally Posted by surjective
Furthermore, I was wondering if the intended could be shown via induction? Is that possible?
Notice that the idea for a proof I gave uses an inductive argument.

Now, just remember that for $f,g\in L^1(\mathbb{R} )$ we define $f\ast g(x)= \int_{\mathbb{R} } f(x-t)g(t)dt = \int_{ \mathbb{R} } g(x-t)f(t)dt = g\ast f (x)$ where the middle equality is by the change of variable theorem. With this definition in hand, and letting $f$ be as I defined it in the previous post we get:

$f\ast N_{m-1}(x) = \int_{ \mathbb{R} } N_{m-1}(x-t)f(t)dt = \int_{0}^{1} N_{m-1}(x-t)dt = N_m(x)$.

Now, you're asked to prove that $\int_{ \mathbb{R} } f\ast N_{m-1} =1$ , but just note:

$\int_{\mathbb{R} } f\ast N_{m-1} = \int_ {\mathbb{R} } \left(\int_{\mathbb{R} } f(x-t)N_{m-1}(t)dt\right) dx$ $= \int_{\mathbb{R} } \left( \int_{\mathbb{R} } f(x-t)N_{m-1}(t)dx\right) dt = \left( \int_{\mathbb{R} } f(y)dy \right) \left( \int_{\mathbb{R} } N_{m-1}(t)dt \right) =1
$

Where we use Fubini (the fact that the theorem applies is left to you) for the second equality and our induction hypothesis for the last.

9. ## Property

Thanks for the response.

It seems that you are well versed in this subject. I have posted another questions elsewhere about the centered B-spline. Would it be possible for you to have a look at it?