Does the map $<f, g>$ $=$ $\int _0^1\:\left(\left(f\left(x\right)-\frac{d}{dx}f\left(x\right)\right)\left(g\left(x\right)-\frac{d}{dx}g\left(x\right)\right)\right)dx$ define an inner product on the set of all polynomial functions of order less than or equal to $n$?

Although I felt that this map was symmetric and bilinear by the properties of definite integrals, I did not think it was positive definite. This was because, say, $<f, f>$ would become $\int _0^1\:\left(\left(f\left(x\right)-\frac{d}{dx}f\left(x\right)\right)^2\right)dx$, which has value greater than zero for values other than the zero function. For example, if $f(x)$ was $2x$, then $f'(x)$ would be $2$ and the value of the above definite integral would be $4/3$. This is why I don't think this space defines an inner product.

So what I am worried about is that I am misunderstanding how the concept of zero vectors with regards to inner products work as this seems to obvious.

If anyone can verify what I have done or tell me how I went wrong, I would greatly appreciate it!

  • 1
    $\begingroup$ Positive definite means exactly that $\langle f,f\rangle\geq 0$ with equality if and only if $f=0$ for all $f$ in your vector space. Your bilinear form has this property, so it should be an inner product as long as you're considering polynomials over $\mathbb{R}$. $\endgroup$ – Melody Nov 15 '18 at 5:11

We have $<f, f>=\int _0^1\:\left(f\left(x\right)-f'\left(x\right)\right)^2dx \ge 0$ and

$<f, f>=0 \iff f=f'$ on $[0,1] $.

Since $f$ is a polynomial, we get

$<f, f>=0 \iff f=0$ on $[0,1] $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.