Question 12.8: Let us denote ej(x) = exp(2jπιx) (j ∈ Z), D^α the derivation......

a) Let us us start from the sequence of functions \left\{f_{m, k+2}(x)\right\}_{m \in \mathbf{N} } (k from (12.23)), where

f_{m, k+2}(x)=\overset{m}{\underset{0≠j=-m}{\sum}} \frac{c_j}{(2 j \pi \imath)^{k+2}} e_j(x) \quad(m \in \mathbf{N} ) .

In view of (12.23), this sequence uniformly converges on every compact set K ⊂ R, hence its limit is a continuous function on R; let us denote it by F_{k+2}.
The distributional derivative of order k + 2 of the function F_{k+2} is

D^{k+2} F_{k+2}(x)=\lim _{m \rightarrow \infty} f_{m, k+2}(x)=f(x)-a_0

In other words, the distribution f, given by (12.24), is the limit of the sequence \left\{a_0+f_{m, k+2}(x)\right\}_{m \in N } \text { in }\mathcal {D} ^{\prime}( \mathbf{R})

b) Since for \alpha \in \mathbf{N} \text { and } m \in \mathbf{Z} _{+} it holds

D^\alpha f_m(x)=\overset{m}{\underset{j=-m}{\sum}}(2 j \pi \imath)^\alpha e_j(x)

part a) implies that the sequence offunctions \left\{D^\alpha f_m\right\}_{m \in N } \text { converges in } \mathcal {D} ^{\prime}( \mathbf{R}) to the distribution D^\alpha f and thus (12.25) holds.

c) From part a) it follows that the sequence \left\{\overset{m}{\underset{j=-m}{\sum}} e_j(x)\right\}_{m \in \mathbf{N} } converges in \mathcal {D} ^{\prime}( \mathbf{R}) to a distribution which we denote by g. Then we have in the sense of \mathcal {D} ^{\prime}( \mathbf{R}):

\left(1-e_1(x)\right) g(x)=\lim _{m \rightarrow \infty}\left(1-e_1(x)\right) \overset{m}{\underset{j=-m}{\sum}} e_j(x)=\lim _{m \rightarrow \infty}\left(e_{-m}(x)-e_{m+1}(x)\right)        (12.26)

Let us prove next that the last limit is equal to 0 in \mathcal {D} ^{\prime}( \mathbf{R}) . To that end, let us analyze the difference e_{-m}(x)-e_{m+1}(x) :

Thus we have

\left|\left\langle e_{-m}(x)-e_{m+1}(x), \varphi(x)\right\rangle\right| \leq \frac{1}{2 \pi m} \int_{ \mathbf{R} }\left|\varphi^{\prime}(x)\right| d x+\frac{1}{2 \pi(m+1)} \int_{ \mathbf{R} }\left|\varphi^{\prime}(x)\right| d x \leq \frac{C}{m}

for some constant C=C(\varphi) . \text { The last right-hand side tends to } 0 \text { as } m \rightarrow \infty, which implies that the right-hand side of (12.26) tends to zero in \mathcal {D} ^{\prime}( \mathbf{R}) as m →∞. Thus we obtained that the sought after distribution g is the solution of the equation

\left(1-e^{2 \pi \imath x}\right) \cdot g(x)=0        (12.27)

For |x|<m \text {, the solutions of the equation } e^{-2 \pi \imath x}=1 are the integers j such that |j| < m. Example 12.7 tells us that the solutions of (12.27) in \mathcal {D} ^{\prime}(-m, m) are exactly the solutions of the following equation

\left(\overset{m-1}{\underset{j=-m+1}{\prod}}(x-j)\right) \cdot g(x)=0        (12.28)

The same example gives us the solution of (12.28) in \mathcal {D} ^{\prime}(-m, m)

g(x)=\overset{m-1}{\underset{j=-m+1}{\sum}} A_j \delta_j(x)

while in \mathcal {D} ^{\prime}( \mathbf{R}) the solution of (12.27) is

g(x)=\underset{j∈\mathbf{Z}}{\sum} A_j \delta_j(x) .

We next show that all constants A_i, j \in \mathbf{Z},are equal to a single constant C.
To that end, note that for a test function \varphi_m such that

\operatorname{supp} \varphi_m \subset(m-2 / 3, m+2 / 3) \text { and } \varphi(x)=1

on the interval (m – 1/3, m + 1/3), it holds

\left\langle g, \varphi_m\right\rangle=\left\langle\underset{j∈\mathbf{Z}}{\sum} A_j \delta_j \varphi_m\right\rangle=\left\langle A_m \delta_m, \varphi_m\right\rangle=A_m        (12.29)

From the other hand, g is 1-periodic, i.e.,

\langle g, \varphi\rangle=\langle g(x), \varphi(x-1)\rangle

for every test function φ But then it follows from (12.29):

A_m=A_{m-1}=C \quad \text { for every } \quad m \in \mathbf{Z} \text {. }

So we get

g(x)=C \cdot \underset{j∈\mathbf{Z}}{\sum} \delta_j(x)