# Gram-Schmidt Orthogonalization for subspace of $L^2$

I am a little stuck on the following problem:

By using the Gram-Schmidt Orthogonalization, find an orthonormal basis for the subspace of $L^2[0,1]$ spanned by $1,x, x^2, x^3$.

OK, so I have defined:

$$e_1 = 1$$

I would then assume that we proceed as follows:

$$e_2 = x - \frac{\langle x,1 \rangle}{\langle 1,1 \rangle} \cdot 1$$

And we have:

$$\langle x,1 \rangle = \int_{0}^{1}x dx = \frac{1}{2}$$

And:

$$\langle 1,1 \rangle = 1$$

So:

$$e_2 = x - \frac{1}{2}$$

But already at this point, my answer is wrong. According to my book, the correct answer here should be:

$$e_2 = \sqrt{12}(x - \frac{1}{2})$$

And this obviously makes all my subsequent answers also incorrect.

So what is it I'm doing wrong here? I simply don't see where the $\sqrt{12}$ term comes from. Any help would be greatly appreciated!

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Hint: What is $\Vert x - \frac{1}{2} \Vert$? –  Alexander Thumm Sep 24 '12 at 9:53
It is required so that $\Vert e_2 \Vert = 1$ –  Tpofofn Sep 24 '12 at 9:57
Ah, I think I get it. We have: $\| x - \frac{1}{2} \| = \sqrt{\int_{0}^{1} (x - \frac{1}{2})^2 dx} = \frac{1}{\sqrt{12}}$. And from there the answer follows. Isn't this correct? –  Kristian Sep 24 '12 at 10:08
Yes it does, @Kristian. The Gram-Schmidt process requires to normalize every vector we get after you've done what you did to get $\,e_2\,$...that's all you were missing. –  DonAntonio Sep 24 '12 at 10:46
Great! Thanks a lot for the help, everyone! –  Kristian Sep 24 '12 at 11:03