I am studying the article "An Extension Problem Related to the Fractional Laplacian" by L. Caffarelli and L. Silvestre. link. At page 2, for a function $f\colon\mathbb{R}^n\to\mathbb{R}$, we consider the extension $u\colon\mathbb{R}^n\times(0,\infty)\to\mathbb{R}$ that satisfies the equation: $$ u(x,0)=f(x)$$ $$\Delta_xu+\frac{a}{y}u_y+u_{yy}=0,$$ I can't prove the following equality: $$ -\lim_{y\to 0^+}y^au_y(x,y)=\frac{1}{1-a}\lim_{y\to0}\frac{u(x,y)-u(x,0)}{y^{1-a}},$$ i have no idea on how to proceed. Can someone help me please ?
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1$\begingroup$ If $a<1$ and the limit of the left exists, then this follows by l'Hospital's rule. $\endgroup$– Iosif PinelisCommented Nov 4, 2020 at 17:26
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$\begingroup$ Sorry, i have dropped the hypothesis $a\in(-1,1)$. $\endgroup$– inocCommented Nov 4, 2020 at 17:35
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$\begingroup$ If i apply l'Hospital's rule, i have that: $$ \frac{1}{1-a}\lim_{y\to0}\frac{u(x,y)-u(x,0)}{y^{1-a}}=\frac{1}{(1-a)^2}\lim_{y\to0}y^au_y(x,y),$$that is: $$ \lim_{y\to0}\frac{u(x,y)-u(x,0)}{y^{1-a}}=\lim_{y\to0}y^au_y(x,y),$$ or there is a mistake in may computation? Moreover doesn't appear the minus sign. Then i obtain $y\to0^+$ in the right limit by continuity? $\endgroup$– inocCommented Nov 4, 2020 at 17:46
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$\begingroup$ Yes, it appears that the constant factor in the equality in question in your post was computed incorrectly. $\endgroup$– Iosif PinelisCommented Nov 4, 2020 at 18:43
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$\begingroup$ That's not a general identity which holds for all $u$. Take for example $u(x, y)=y^{1-a}$, you'll see that the LHS equals $(a-1)$ while the RHS equals $\frac{1}{1-a}$. For that identity to hold, $u$ must satisfy the given PDE. In any case, from the way the author phrase the result, I understand that they are going to prove the identity later in the paper. $\endgroup$– Giuseppe NegroCommented Nov 4, 2020 at 19:07
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