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Let $\Phi(t)=a+\frac{1}{2}(1+t)(b-a)$. Then $\Phi'(t)=\frac{b-a}{2}$.

Now let's transform the following $L^2$-norm from the interval $[-1,1]$ to the interval $[a,b]$, trough the substitution: $$ \begin{align*} x&=\Phi(t) \\ dx&=\Phi'(t)\,dt \quad\Longleftrightarrow \quad dt =\frac{1}{\Phi'(t)}\, dx =\frac{2}{b-a}\, dx \end{align*} $$ So, $$ \begin{align*} ||{(f\circ\Phi)}||_{L^2_{[-1,1]}} &=\sqrt{\int_{-1}^1|f(\Phi(t))|^2\, dt}\\ &=\sqrt{\int_{\Phi(-1)=a}^{\Phi(1)=b}|f(x)|^2\frac{2}{b-a}\, dx}\\ &=\sqrt{\frac{2}{b-a}}\sqrt{\int_{a}^{b}|f(x)|^2\, dx}\\ &=\sqrt{\frac{2}{b-a}}||{f}||_{L^2_{[a,b]}}. \end{align*} $$

However, the solution in my book says $||{(f\circ\Phi)}||_{L^2_{[-1,1]}}=\sqrt{|b-a|}\cdot||{f}||_{L^2_{[a,b]}}$. What have I done wrong?

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    Probably the book uses $$\lVert f\rVert_{L^2[c,d]}^2 = \frac{1}{d-c}\int_c^d \lvert f(u)\rvert^2\,du,$$ normalising the measure of the space to $1$. – 2017-01-07
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    Err, no. That would make composition with $\Phi$ an isometry. $$\lVert f\rVert_{L^2[c,d]} = \frac{1}{d-c}\sqrt{\int_c^d \lvert f(u)\rvert^2\,du}$$ would do it. But that's a weird normalisation. – 2017-01-07
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    The book defines the L^2 norm for an arbitrary interval without normalisation. – 2017-01-07
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    Well, take $f \equiv 1$. Without normalising factors, $\lVert 1\rVert_{L^2[c,d]} = \sqrt{d-c}$. So if there is a constant $K$ such that $\lVert f\circ \Phi\rVert_{L^2[-1,1]} = K\cdot \lVert f\rVert_{L^2[a,b]}$ for all $f\in L^2[a,b]$, then $\sqrt{2} = K\cdot \sqrt{b-a}$, so only yours has a chance to be right. (It is right, then.) So whoever wrote the solutions dun goofed. Happens all the time. – 2017-01-07

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