Orthogonal basis of specific span in the continuous function space

Question

Consider the function space of continuous functions defined on $x\in(0,1)$ with inner product $$\langle u,v\rangle =\int^{1}_{0}u(x)v(x)\,{\rm d}x.$$ Find any orthogonal basis for span $\{1,x\}$.

My attempt: The span $\{1,x\}$ is basically the set of all linear combinations of the two functions $f(x)=1$ and $g(x)=x$, for example $h(x)=\beta\cdot1+\alpha\cdot x$, where $\alpha,\beta$ are arbitrary scalars.

My task is to find an orthogonal basis, that is, a set of functions that can express in a linear combination any function in my span and also satisfy orthogonality, that is, $\langle u,v\rangle =0$.

So, let $\{a,b\}$ be my orthogonal basis. Since my span is simple enough, let $a=1$ and $b=c_1(1)+c_2(x)$, where $c_1,c_2$ are scalars that I want to determine.

Now I test orthogonality: $$\langle a,b\rangle =\int^{1}_{0}ab\,{\rm d}x=\int^{1}_{0}(1)(c_1+c_2x)\,{\rm d}x=c_1+c_2/2=0,$$ and so we have $c_2=2, c_1=-1$.

Thus $b=-1(1)+2(x)=2x-1$. That gives me $\{1,2x-1\}$ as the orthogonal basis of the span $\{1,x\}$.

Is my solution wrong? Have I done anything invalid? I would appreciate any feedback.

Answer 1

You've got it written up well. I'm not the best, but since you want someone to check the words you are using:

• "the span $\{ 1,x\}$ is basically the set of..." well yes, because the space $\operatorname{span}\{1,x\}$ is by definition exactly the set of... .
• "and so we have $c2=2,c1=−1$" Strictly speaking, this doesn't follow, because there are other solutions, as long as $c_1 = -c_2/2$, like $c_1 = \pi, c_2 = -2\pi$, and similarly,
• "this gives me ... as the orthogonal basis" there are many other choices for orthogonal bases. Even the description from the previous point doesn't describe them all.

Also, you should be using \langle a,b\rangle to type $\langle a,b\rangle$ instead of <a,b> to type $<a,b>$.