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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!

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    $\begingroup$ 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
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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] $.

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