# Proof that the harmonic function defined by the Poisson integral is continuous on the boundary of a ball.

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 $$|\phi(y)| on $$\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

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

• Ah how obvious! I am kicking myself. Thank you for your answer Kavi. – valcofadden Jul 20 at 12:28