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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 :$.

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  • $\begingroup$ What is $\Pi ^1$ exactly? $\endgroup$
    – Git Gud
    Jun 15 '13 at 16:27
  • $\begingroup$ The vector space of the polynomials with grade less than or equal to 1. Is not the right notation? $\endgroup$
    – Alfageme
    Jun 15 '13 at 16:30
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    $\begingroup$ I've never seen it before, but that doesn't mean anything. Thanks for clarifying. $\endgroup$
    – Git Gud
    Jun 15 '13 at 16:46
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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.

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  • $\begingroup$ 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! $\endgroup$
    – Alfageme
    Jun 15 '13 at 16:50
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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.

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  • $\begingroup$ I avoided splitting orthonormal and orthogonal as before. $\endgroup$
    – Avitus
    Jun 15 '13 at 16:46

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