Here I am looking at the proof of theorem 2 below
Here I have the following difficulties:
1) In the last two lines, the exponent changes from $\frac{n-1}{2}$ to $\frac{n-3}{2}$, why? Could anyone explain?
2) To show (iii), i.e. check I.C. satisfied. I follow identities in lemma2 (ii) and (iii).(See picture below)
But it seems I am lost and have messed up my calculation again. Could anyone help?
On
The set on which the integral is evaluated also changes. Here he's using polar coordinates to make the calculation work: $$ \int_{B(x,t)} f(y) dy = \int_0^t \int_{\partial B(x,s)} f(z) d\sigma(z) ds $$ where $d\sigma$ is the surface measure on the sphere. Hence the fundamental theorem of calculus says that $$ \frac{d}{dt} \int_{B(x,t)} f(y) dy = \int_{\partial B(x,t)} f(z) d\sigma(z). $$
In the book he is taking one copy of the operator $t^{-1} \frac{d}{dt}$ and applying it as above, which then decreases the power by one and results in the term $$ \frac{1}{t} \int_{\partial B(x,t)} \Delta h . $$
For one, $$\left(\frac{1}{t}\frac{d}{dt}\right)^{\frac{n-1}{2}} = \left(\frac{1}{t}\frac{d}{dt}\right)^{\frac{n-3}{2}}\left(\frac{1}{t}\frac{d}{dt}\right)$$ Then note that the quantity $\left(\frac{1}{t}\frac{d}{dt}\right)$ appears to get tucked into the integral. This is why on the second line in question you have a $\frac{1}{t}$ outside the integral. The infinitesimal also changes from $dy$ to $dS$ along with the integral region. Using the formula $$\frac{d}{dt}\left(\int_{B(x_0,t)}f dx\right)=\int_{\partial B(x_0,t)} f dS$$ you mentioned as appearing on appendix pg. 628 of Evans, the result follows.