In vector space of all real polynomials with inner product $(x,y) = \sum_0^1x(t)y(t)dt$. $x_n(t) = t^n$ for $n = 0, 1, \dots $.
Show that functions: $y_0(t) = 1, $
$y_1(t) = \sqrt(3)(2t-1), $
$y_2(t)= \sqrt(5)(6t^2 -6t+1)$
are an orthonormal pair that span the same subspace as $\{x_0,x_1x_2\}$.
Plan of attack: (1) take inner product of $y_0,y_1$, $y_1,y_2$, and $y_2,y_3$. All 3 inner products should equal 0.
Question: how do I show that the functions $y_0,y_1,y_2$ are orthoNORMAL?
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$\begingroup$I think you mean the inner product on the space is $\langle x,y\rangle:=\int_{0}^1 x(t)y(t)\,dt$.
To show the set you stated is an orthonormal set, you want to show that it is an orthogonal set and each member of the set has unit norm. You can show orthogonality by showing that the inner product of each (distinct) pair is zero.
To show that each member as unit norm, recall that the norm induced by the inner product is $\|f\|:=\langle f,f\rangle^{1/2}=\sqrt{\int_0^1 f^2(t)\,dt}$, so you want $\|y\|=\sqrt{\int_0^1 y_i^2(t)\,dt}=1$ for $i=0,1,2$.
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