Show that a finite set $B = \left\{1, x, x^2 \right\}$ is an orthonormal system with respect to the inner product $$\left \langle f, g \right \rangle = \int _{-1}^1\:f\left(t\right)\cdot g\left(t\right) \,\rm{d}t$$ for any $f, g \in L^2 \left[-1, 1 \right]$. Hint: evaluate $\left \langle 1, x \right \rangle, \left \langle 1, x^2 \right \rangle, \left \langle x, x^2 \right \rangle$.
As far as I'm concerned, these are the indicators for orthogonality (if they're all zero then the set is orthogonal). My results are consecutively $2x$, $2x^2$, and $2x^3$.
I don't understand – they do equal zero only if $x = 0$. Does that mean that the set is not orthogonal – and therefore cannot be orthonormal? How do I proceed with this problem?
Best Answer
$\int_{a}^b f(x)\,dx$ is not a function of $x$, but a number (like, $37$, $92$, $\pi$...). Said number is not to be confused with the function $F(t)= \int_c^t f(x)\,dx$, although the latter is often useful because $\int_a^b f(x)\,dx=F(b)-F(a)$ and $F$ may be recovered via the relation $F'=f$ if $f$ is continuous.
As for the other observation, yes, strictly speaking orthogonal means $\langle u,v\rangle$ for all $u\ne v$, whereas orthonormal means orthogonal + $\langle u,u\rangle=1$ for all $u$.
It is worth mentioning that the set $\{1,x,x^2\}$ is not orthogonal, though.
Added: Now that I think about it, perhaps you did not realise that $x$ does not stand for some number $x$, but rather for the function $f(x)=x$. Likewise, $x^2$ stands for the function $x\mapsto x^2$.