# Inner Product, Definite Integral

Does the map $$$$ $$=$$ $$\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, $$$$ would become $$\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!

• 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 $\mathbb{R}$. – Melody Nov 15 '18 at 5:11

We have $$=\int _0^1\:\left(f\left(x\right)-f'\left(x\right)\right)^2dx \ge 0$$ and
$$=0 \iff f=f'$$ on $$[0,1]$$.
Since $$f$$ is a polynomial, we get
$$=0 \iff f=0$$ on $$[0,1]$$.