# A question about inner product and Gram-Schmidt process

Let there be the following bilinear form: $\int_0^1f(x)g(x)x\,dx$, which acts on the polynomials with degree $\leq2$. I needed to prove it's an inner product and then find an orthonormal basis. I needed to use Gram-Schmidt proccess.

So, when I make the vectors I find to be of length one, what's the inner product I use? Lets say some vector for basis if $h$, then the normal of the vectors is $\sqrt{\int_0^1h(x)h(x)x\,dx}$, or is it the 'standard' inner product $\sqrt{\int_0^1h(x)h(x)\,dx}$? In other words, when a basis is orthnormal, it is orthogonal and of length one with accordance to some specific inner product and not necessarily others?

The definite integral is your Inner Product. Now, in order to build all polynomials that have degree less or equal than 2 you need a basis like $x^0$, $x$ and $x^2$.

$$B_{x^0}^N=1,\text{ as }\int\limits_0^1~1\cdot1=1$$

$$B_{x^1}=x-\frac{1}{2}~~=~~x-\frac{\langle 1,x\rangle}{\langle 1,1\rangle}\cdot1$$

Now, the norm of $x-\frac{1}{2}$ is $\sqrt{\frac{1}{12}}$, given by the integral you posted.

$$B_{x^1}^N=2\sqrt{3}·x - \sqrt{3}$$

I'll let you guess the vector that spans the $x^2$ part.

• My inner products is $\int_0^1f(x)g(x)xdx$. So I normalize vectors using that inner product? Notice it's -not- the standard inner products, and that the polynomials are also multiplied by $x$.
– Jim
Feb 1, 2014 at 6:05