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[SOLVED] Euclidean Space of Polynomials

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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 \(U=<x>\).

My Answer:

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
Jan 25, 2013
236
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
Feb 5, 2012
1,621
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
Jan 25, 2013
236
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.