Is the Space of Real Polynomials of Degree ≤ n a Euclidean Space?

[Sudharaka]

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?

 

[johng1]

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]

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 \(a\).

 

[johng1]

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.

 

[blue_raver22]

If so, then we can check that it is indeed positive and symmetric, thus satisfying the criteria for an Euclidean space.

For part b), we have \(f(x)=1\) and \(U=<x>\). So, we can write \(f(x)=cx\) for some constant \(c\). The distance between two vectors \(f(x)\) and \(g(x)\) in an Euclidean space is given by the norm \(\left\lVert f(x)-g(x)\right\rVert=\sqrt{\left(f(x),\,f(x)\right)-2\left(f(x),\,g(x)\right)+\left(g(x),\,g(x)\right)}\).

In this case, we have \(g(x)=cx\) and using the bilinear function given in the question, we get

\[\left(f(x),\,f(x)\right)=\int_{0}^{1}1\cdot1\,dx=\frac{1}{2}\]

\[\left(f(x),\,g(x)\right)=\int_{0}^{1}1\cdot cx\,dx=\frac{c}{2}\]

\[\left(g(x),\,g(x)\right)=\int_{0}^{1}cx\cdot cx\,dx=\frac{c^2}{3}\]

Hence, the distance from \(f(x)=1\) to the linear span \(U=<x>\) is given by

\[\left\lVert f(x)-g(x)\right\rVert=\sqrt{\frac{1}{2}-2\cdot\frac{c}{2}+\frac{c^2}{3}}=\sqrt{\frac{1}{12}-\frac{c}{2}+\frac{c^2}{3}}\]