Affine normalization isometry for profile coordinates

The affine map $T:(0,1)\to I$, $T(\xi )=a+\mathrm{\ell }\xi$, is a $C^1$ diffeomorphism with derivative $\mathrm{\ell }>0$. If $q_1,q_2$ represent the same element of $H_I$, the measurable set $N=\{z\in I:q_1(z)\mathrm{\ne }{q}_2(z)\}$ is null. Apply [F2] to ${\mathbf{1}}_N$ and $T$: $0={\lambda }_I(N)=\mathrm{\ell }{\lambda }_{(0,1)}(T^{-1}N)$, so $T^{-1}N$ is null. Therefore the displayed assignment is independent of representatives, and its pointwise formula is linear on representative functions.

For a representative $q$ of an element of $H_I$, apply [F2] to the nonnegative function $z\mapsto \mathrm{\mid }q(z){\mathrm{\mid }}^2$. The formula gives ${\int }_I\mathrm{\mid }q(z){\mathrm{\mid }}^{2}dz=\mathrm{\ell }{\int }_0^1\mathrm{\mid }q(a+\mathrm{\ell }\xi ){\mathrm{\mid }}^{2}d\xi =\mathrm{\parallel }{U}_{a,\mathrm{\ell }}[q]{\mathrm{\parallel }}_{H_0}^2$. The right-hand integral is finite by $q\in H_I$, so the image is in $H_0$, $U_{a,\mathrm{\ell }}$ is a well-defined linear map, and the norm is preserved.

The inverse affine map $T^{-1}:I\to (0,1)$ is also a $C^1$ diffeomorphism. The same null-set argument makes $V_{a,\mathrm{\ell }}$ independent of the representative, and change of variables for $T^{-1}$, whose derivative is $1\mathrm{/}\mathrm{\ell }$, gives $\mathrm{\parallel }{V}_{a,\mathrm{\ell }}[g]{\mathrm{\parallel }}_{H_I}^2={\mathrm{\ell }}^{-1}{\int }_I\mathrm{\mid }g((z-a)\mathrm{/}\mathrm{\ell }){\mathrm{\mid }}^{2}dz={\int }_0^1\mathrm{\mid }g(\xi ){\mathrm{\mid }}^{2}d\xi$. Direct substitution gives $V_{a,\mathrm{\ell }}{U}_{a,\mathrm{\ell }}[q]=[q]$ and $U_{a,\mathrm{\ell }}{V}_{a,\mathrm{\ell }}[g]=[g]$; hence $V_{a,\mathrm{\ell }}$ is the inverse and $U_{a,\mathrm{\ell }}$ is bijective.