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For square-integrable periodic boundary data and the Fourier coefficients used in the Poisson kernel for the unit disk, the displayed identity gives the squared area norm of its Poisson extension $u$. At fixed radius $r$, Parseval's identity givesMultiply by the polar coordinates area element factor $rdr$ and integrate from zero to one. The constant term contributes $\pi |a_0{|}^2/4$, and ${\int }_0^1{r}^{2n+1}dr=1/[2(n+1)]$. Interchanging the nonnegative sum and radial integral is justified by the monotone convergence theorem. The resulting factor $1/(n+1)$ shows how higher angular frequencies contribute less to the interior area norm.

For fixed $r<1$, $\sum r^n\cos (n\theta )$ converges absolutely and uniformly. Thus for integrable $f$, one may integrate the kernel series term by term. The cosine addition formula givesSubstitute this into the defining integral. The constant kernel term gives $\frac{1}{2\pi }\int f=a_0/2$; the remaining terms givewhereThese are the same Fourier coefficients as in part (a). The kernel $2K$ is the Poisson kernel for the unit disk; the factors $r^n$ suppress the higher angular frequencies inside the disk.

For $0\le r<1$, this Poisson kernel for the unit disk has the uniformly convergent Fourier seriesThe identity follows by taking the real part of the geometric series $1+2{\sum }_{n\ge 1}(r{e}^{i\theta }{)}^n$. For integrable $2\pi$-periodic boundary data $f$, its Poisson extension iswhere $a_n,b_n$ are the usual Fourier coefficients. Each term is a harmonic function in the open disk. For continuous boundary data, the extension solves the Dirichlet problem with those boundary values.