I am reading the text Elliptic Partial Differential Equations of Second Order by D.Gilbard and N.Trudinger, I am struggling with a particular proof they have given. (p.21)

In the text they are seeking to prove that the function defined on the ball $B=B_R(0)$, by

\begin{equation} u(x)= \begin{cases} \int_\limits{{\partial B}}K(x,y)\phi(y)ds_y \qquad x \in B\\ \phi(y) \qquad \qquad \qquad \qquad \qquad \quad x \in \partial B \end{cases} \end{equation} is continuous on $\partial B$. Here the integral is done over the $y$ variable, $\phi$ is a continuous function on $\partial B$ and $K(x,y)$ is the Poisson kernel;

$$K(x,y)=\frac{R^2-|x|^2}{s_nR|x-y|^n}\qquad x \in B \quad y \in \partial B$$ $s_n$ is the surface area of the unit $n$ sphere.

The proof goes like this;

Let $x_0 \in \partial B$. By continuity of $\phi$ for $\epsilon>0$ we can take $\delta>0$ such that $$|x-x_0|<\delta \implies |\phi(x)-\phi(x_0)|< \epsilon$$as usual. Now if we let $|x-x_0| <\frac{\delta}{2}$ and assume that $|\phi(y)|<M$ on $\partial B$, the author derives the estimate $$|u(x)-u(x_0)|\leq \epsilon + \frac{2M(R^2-|x^2|)R^{n-2}}{\left(\frac{\delta}{2}\right)^n} < 2\epsilon$$

Where the second inequality above follows by taking $|x-x_0|$ sufficiently small.

What I am struggling to understand is how this second inequality follows. By taking $|x-x_0|$ to be sufficiently small I assume what they mean is to introduce $\delta'$ such that $|x-x_0|<\delta'<\frac{\delta}{2}$ so the above estimate holds, and make $\delta'$ smaller as this will squeeze the size of $|x-x_0$| down. This would make sense as then $(2\epsilon,\delta')$ would serve as the required epsilon and delta to give the continuity of $u$.

However I just can't see how they've gotten the upper bound of $2\epsilon$ above by squeezing $|x-x_0|$?

Any help would be appreciated.


1 Answer 1


$x_0 \in \partial B$ so $|x_0| =R$. Hence $R^{2}-|x|^{2}$ can be made as small as wish by making $|x-x_0|$ small.

  • $\begingroup$ Ah how obvious! I am kicking myself. Thank you for your answer Kavi. $\endgroup$ Jul 20, 2019 at 12:28

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