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In Evans' PDE book he gives the following theorem. Assume $u$ is harmonic in $U$. Then, $$ |D^{\alpha}u(x_0) | \le \frac{C_k}{r^{n+k}}||u||_{L^1(B(x_0,r))}$$

When asking my professor for some motivation for the above inequality she had she jotted down the following, $$D u(x) = \lim_{h \to 0} \frac{u(x+h) - u(x) }{h}$$ $$ = \lim_{r \to 0} \frac{u(x+r) - u(x)}{r}$$ and so for tiny $r(x)$, we have that $D u(x) \sim \frac{u(x)}{r}$, assuming $u : \mathbb{R} \to \mathbb{R}$.

However, $D u(x) \sim \frac{u(x)}{r}$ does not make much sense to me. Can anyone explain what was meant by this?

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I haven't found this explanation illuminating myself: in the estimate, $r$ is the radius of the ball, which is a fixed number. So $r\to 0$ looks strange.

I prefer to write this estimate in equivalent form $$|D^{\alpha}u(x_0) | \le \frac{C_k}{r^{k}} |u|_{B(x_0,r)}$$ where $|u|_E$ means the average of $|u|$ on the set $E$.

In this form, the appearance of $r^k$ in the denominator makes sense in terms of scaling. If the whole picture is magnified by some factor $\lambda$, the derivatives get multiplied by the factor of $\lambda^{-k}$ (chain rule), while the averages stay the same.

The above paragraph allows one to reduce the problem to $$|D^{\alpha}u(x_0) | \le {C_k} |u|_{B(x_0,1)}$$

At this stage, I would observe that the average over $B(x_0,r)$ controls the spherical average over $\partial B(x_0,r)$ for some $r\in (1/2,1)$. This is just Fubini's theorem, not specific to harmonic functions.

Finally, spherical averages control the derivative by the Poisson's formula. This is the only step where harmonicity matters.

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