Orthonormalize the set of functions {1,x}

I'm having trouble while doing this exercise, it says:

In the vector space of the continuous functions in [0,1] with the inner product : $$\langle f,g \rangle = \int_{0}^{1}f(x)g(x)dx$$ a) Orthonormalize the set of functions $\left\{1,x\right\}$

So I assumed $B$ as a base of the vector space $\prod^{1}$ (the canonical base) $B = \left\{(1,0),(0,1)\right\}$ and applied Gram-Schmidt method to orthonormalize the functions.

The first one it's already normalized for that inner product $\vec{u_{1}} = \frac{\vec{v_{1}}}{\|\vec{v_{1}}\|} = (1,0)$

So, for the second one $\vec{u_{2}} = \frac{\vec{v_{2}}-\langle \vec{v_{2}},\vec{u_{1}}\rangle \vec{u_{1}}}{\|\vec{v_{2}}-\langle \vec{v_{2}},\vec{u_{1}}\rangle \vec{u_{1}}\|} = \frac{(0,1)-\left(\frac{1}{2},0\right)}{\|\left(-\frac{1}{2},1\right)\|}=(-6,12)$

But then $\|\vec{u_{2}}\|=\sqrt{\langle \vec{u_{2}},\vec{u_{2}}\rangle} = \sqrt{12} \neq 1$ ; so $\vec{u_{2}}$ is not normal.

I don't know whether I'm all wrong assuming the base or doing it wrong with the Gram-Schmidt method, but I've been stuck for more than an hour with this :$. • What is$\Pi ^1$exactly? Jun 15 '13 at 16:27 • The vector space of the polynomials with grade less than or equal to 1. Is not the right notation? Jun 15 '13 at 16:30 • I've never seen it before, but that doesn't mean anything. Thanks for clarifying. Jun 15 '13 at 16:46 2 Answers You don't need to use the standard basis, you already have a basis of$\{1,x\}$for the space you're interested in, namely$Span(\{1,x\})$. Use Gram-Schmidt on those vectors directly. As you point out$\langle 1,1\rangle=1$already, so you have one orthonormal basis element already. For the other one, take$x-\frac{\langle x,1\rangle}{\langle 1,1\rangle}1=x-\langle x,1\rangle=x-\int_0^1xdx=x-0.5x^2|_0^1=x-\frac{1}{2}$. Now$x-\frac{1}{2}$is orthogonal to$1$, and you need to normalize it to make it length$1$. We calculate$\langle x-0.5,x-0.5\rangle=\int_0^1 (x-0.5)^2dx=\int_0^1x^2-x+0.25dx=\frac{1}{3}x^3-\frac{1}{2}x^2+0.25x|_0^1=\frac{1}{3}-\frac{1}{2}+\frac{1}{4}=\frac{1}{12}$. Putting it all together,$\{1,x\sqrt{12}-\sqrt{3}\}$is an orthonormal basis for the desired space. • Ok, I get my error, I multiplied$(-\frac{1}{2},1)$by 12 instead of$\sqrt{12}$and get messed up at the end. Thanks! Jun 15 '13 at 16:50 If I correctly understood your question, you want to find a ortonormal basis of$V$, where$V=Span\{1,x\}$w.r.t the given scalar product. In other words, you want to begin by correcting through normalization the fact that $$\langle 1, x\rangle=\int_0^1 x dx=\frac{1}{2}$$. If you search for an orthonormal basis, then you need to introduce at first$\{1,x-\frac{1}{2}\}$as the computation above imply orthogonality. All what remains is to normalize the element$x-\frac{1}{2}$by computing at first $$\|x-\frac{1}{2}\|^2=\langle x-\frac{1}{2} , x-\frac{1}{2} \rangle=\frac{1}{12}$$. Then$\{1, 2\sqrt{3}(x-\frac{1}{2})\}\$ is orthonormal.

• I avoided splitting orthonormal and orthogonal as before. Jun 15 '13 at 16:46