# Gradient Estimate for Logarithmic Cutoff Function

In Notes 4 on regularity of harmonic maps, T. Tao asserts the following lemma, which in short, allows us to localize a map $u:\Omega\rightarrow S^{m-1}$ (here, $\Omega\subset\mathbb{R}^{2}$ is a bounded open set) with finite energy (i.e. $\int|\nabla u|^{2}<\infty$) around a point in $\Omega$ so that the energy of the localization is arbitrarily small.

Lemma. Let $x_{0}\in\Omega$, $u:\Omega\rightarrow S^{m-1}$ be a function with finite energy, and let $\delta>0$. Then there exists an $0<r<c_{0}$ and a $C_{c}^{\infty}$ function $\eta$ supported on a ball $B_{c_{0}}(x_{0})\subset\Omega$ such that $\eta(x)=1$ on $B_{r}(x_{0})$, we have the estimates

1. $\int|\nabla(\eta u)|^{2}<\delta$
2. $|\eta(x)|\lesssim\dfrac{\log\frac{2c_{0}}{r+|x-x_{0}|}}{\log\frac{2c_{0}}{r}}$
3. $|\nabla^{j}\eta(x)|\lesssim_{j}\dfrac{1}{\log\frac{2c_{0}}{r}}\dfrac{1}{(r+|x-x_{0}|)^{j+1}}, \quad\forall j\geq 1$

To prove the lemma, we do something as follows. Let $c_{0}$ be such that $B_{c_{0}}(x_{0})\subset\Omega$, and let $0<r\ll c_{0}$ be a small number to be determined later (the implicit constants won't depend on r). Then let $\eta$ is a function which equals $1$ on $B_{r}(x_{0})$, equals $\dfrac{\log\frac{2c_{0}}{r+|x-x_{0}|}}{\log\frac{2c_{0}}{r}}$ on $B_{c_{0}/2}\setminus B_{2r}(x_{0})$, and smoothly interpolates in the regions $B_{2r}(x_{0})\setminus B_{r}(x_{0})$ and $B_{c_{0}}(x_{0})\setminus B_{c_{0}/2}(x_{0})$. From this, Tao claims we get (2) and (3) and then by using the estimate $$\int |\nabla(\eta u)|^{2}\lesssim\int \eta^{2}|\nabla u|^{2}+\int|\nabla \eta|^{2}|u|^{2},$$ we get (1).

My Problem I don't see how we get (1).

I am fine with Tao's claim that the first integral on the RHS is $\lesssim \epsilon$ for $r$ sufficiently small by dominated convergence. But I don't see why the estimate for $\nabla\eta$ gives us that the second integral can be made small by taking $r$ sufficiently small. Since $\eta\equiv 1$ on $B_{r}(x_{0})$, $\nabla\eta=0$ on $B_{r}(x_{0})$. So since $|u|=1$, we have the estimate \begin{align*} \int|\nabla \eta|^{2}|u|^{2}&\lesssim\int_{r<|x-x_{0}|\leq c_{0}}\dfrac{1}{(\log\frac{2c_{0}}{r})^{2}}\dfrac{1}{(r+|x-x_{0}|)^{4}}dx\\ &\lesssim \dfrac{1}{r^{2}(\log\frac{2c_{0}}{r})^{2}}\int_{1<|x-x_{0}|}\dfrac{1}{(1+|x-x_{0}|)^{4}}dx \end{align*} by dilation invariance. The last expression blows up as $r\rightarrow 0$, so is useless.

It seems based on the construction of $\eta$ that we should instead have the estimate

$$|\nabla^{j} \eta(x)|\lesssim\dfrac{1}{\log\frac{2c_{0}}{r}}\dfrac{1}{(r+|x-x_{0}|)^{j}},\quad\forall j\geq 1, \enspace x\in\Omega$$ This estimate together with dilation invariance would then give us $$\int|\nabla \eta|^{2}|u|^{2}\lesssim\dfrac{1}{(\log\frac{2c_{0}}{r})^{2}}\int_{|x-x_{0}|>1}\dfrac{1}{(1+|x-x_{0}|)^{2}}dx,$$ which $\rightarrow 0$, as $r\rightarrow 0$.

The numbered estimate (3) has a typo: the exponent should be $j$, not $j+1$. That is, $$|\nabla^{j}\eta(x)|\lesssim_{j}\dfrac{1}{\log\frac{2c_{0}}{r}}\dfrac{1}{(r+|x-x_{0}|)^{j}}, \quad\forall j\geq 1$$ Indeed, $(\log x)'=x^{-1}$ and so forth. Then after integration in polar coordinates, \begin{align*} \int|\nabla \eta|^{2}|u|^{2}&\lesssim\int_{r<|x-x_{0}|\leq c_{0}}\dfrac{1}{(\log\frac{2c_{0}}{r})^{2}}\dfrac{1}{(r+|x-x_{0}|)^{2}}dx = \dfrac{2\pi \log\frac{c_0}{r}}{(\log\frac{2c_{0}}{r})^{2}} \end{align*} which tends to $0$ as $r\to 0$.
Same in cleaner notation: $$\int_{r<|x|<R}\left|\nabla \frac{\log (R/|x|)}{\log (R/r)}\right|^2 = \frac{1}{\log^2(R/r)}\int_r^R r^{-2} (2\pi r)\,dr = \frac{2\pi}{\log (R/r)}$$
In potential theory, this is how one proves that a singleton in the plane has zero logarithmic capacity. For the same reason, $W^{1,2}$ norm on the plane does not control pointwise values, i.e., $W^{1,2}$ fails to embed into $L^\infty$.