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I have the following problem: Setting, for $\lambda>0$, $x_\lambda\in\mathbb{R}^d$, $$ (g_\lambda u)(x):=\lambda^{-\frac{d-2}{2}}u\left(\frac{x-x_\lambda}{\lambda}\right)\quad e \quad [(g_\lambda)^{-1}u](x):=\lambda^{\frac{d-2}{2}}u(\lambda x+x_\lambda). $$

Show that $\Vert g_\lambda u\Vert_{\dot{H}^1}=\Vert u\Vert_{\dot{H}^1}=\Vert (g_\lambda)^{-1}u\Vert_{\dot{H}^1}$.

But when I use the change of variables $y=\frac{x-x_\lambda}{\lambda}$, I got

$$ \Vert g_\lambda u\Vert_{\dot{H}_x^1}= \int_{\mathbb{R}^d}|\nabla (g_\lambda u)|^2dx=\int_{\mathbb{R}^d}\left|\nabla\left[\lambda^{-\frac{d-2}{2}}u\left(\frac{x-x_\lambda}{\lambda}\right)\right]\right|^2dx$$$$=\lambda^{-(d-2)}\lambda^d\int_{\mathbb{R}^d}|\nabla u(y)|^2dy=\lambda^2\Vert u\Vert_{\dot{H}^1} $$

What am I doing wrong?

Sebastiano
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mshespanha
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    Do you know how the change of variables formula works for the much simpler but closely related integral $\int_{\mathbb R^d}f(\alpha x),dx$ when you substitute $y=\alpha x$? Hint: you need the determinant of a Jacobi matrix. – Kurt G. Mar 07 '23 at 20:10
  • Isnt $\alpha^{-d}$ the determinant of the Jacobi matrix in this case? Because is a diagonal matrix with $\alpha$ in the main diaogonal – mshespanha Mar 08 '23 at 11:22
  • Please figure out yourself if it is $\alpha^d$ or $\alpha^{-d},.$ See here for example how to do it from scratch. Suppose I answer your last question with yes/no. What if I err? Hint is given. Work it out. – Kurt G. Mar 08 '23 at 11:43

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