# [SOLVED]Euclidean Space of Polynomials

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone, Here's a question I encountered and I need your help to solve it.

Question:

Let $$V$$ be the space of real polynomials of degree $$\leq n$$.

a) Check that setting $$\left(f(x),\,g(x)\right)=\int_{0}^{1}f(x)g(x)\,dx$$ turns $$V$$ to a Euclidean space.

b) If $$n=1$$, find the distance from $$f(x)=1$$ to the linear span $$U=<x>$$.

In our notes it's given that an Euclidean space is a pair $$(V,\,f)$$ where $$V$$ is a vector space over $$\mathbb{R}$$ and $$f:V\times V\rightarrow\mathbb{R}$$ is a positive symmetric bilinear function. So therefore I thought that we have to check whether the given bilinear function is positive symmetric. It's clearly symmetric as interchanging $$f$$ and $$g$$ won't matter. But to make it positive we shall find a condition. Let,

$f(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$

$g(x)=b_{0}+b_{1}x\cdots+b_{n}x^{n}$

Then,

$f(x)g(x)=\sum_{k=0}^{n}\left(\sum_{i=0}^{n}a_{i}b_{k-i}\right)x^{k}$

Hence we have,

$\left(f(x),\,g(x)\right)=\int_{0}^{1}f(x)g(x)\,dx>0$

$\Rightarrow \sum_{k=0}^{n}\left(\sum_{i=0}^{n}a_{i}b_{k-i}\right)\frac{1}{k+1}>0$

Is this the condition that we have to obtain in order for $$V$$ to become an Euclidean space?

#### johng

##### Well-known member
MHB Math Helper
Usually, a bilinear function f is positive definite iff f(v,v) is non-negative for all v and f(v,v) > 0 for v not zero. If this is what you meant, your conclusion is certainly true. (Apparently, you think positive means f(u,v) > 0 for all u,v. For this interpretation, your result is clearly false. Furthermore, it's impossible to have such a "positive" bilinear form.)

#### Sudharaka

##### Well-known member
MHB Math Helper
Usually, a bilinear function f is positive definite iff f(v,v) is non-negative for all v and f(v,v) > 0 for v not zero. If this is what you meant, your conclusion is certainly true. (Apparently, you think positive means f(u,v) > 0 for all u,v. For this interpretation, your result is clearly false. Furthermore, it's impossible to have such a "positive" bilinear form.)
Hi johng, Thanks much for the reply. Yeah, I see the mistake in my interpretation of positive definite. So it seems that the result should be,

$\Rightarrow \sum_{k=0}^{n}\left(\sum_{i=0}^{n}a_{i}a_{k-i}\right)\frac{1}{k+1}>0$

Am I correct? I replaced the $$b$$ by $$a$$.

#### johng

##### Well-known member
MHB Math Helper
For a fixed v, you must show (v,v)>=0 and (v,v)=0 only for v=0. So for your particular scalar product, let g be a polynomial of degree at most n. Then

$$\displaystyle (g,g)=\int_0^1g(x)^2\,dx\geq0$$ since the integral of a non-negative function is non-negative.

Also $$\displaystyle \int_0^1g(x)^2\,dx=0\text{ implies }g\equiv0$$ which is true by the fact that if the integral of a non-negative function is 0, then the function must be identically 0.