Séries de Dirichlet.
The following exercise will help you apply and better understand Abel transform. It's also an example of a series of functions that converges uniformly on a subset of \(\mathbb{C}\), but not necessarily normally.
Exercise: Let \((a_n)_{n\in\mathbb{N}}\in \mathbb{C}^{\mathbb{N}}\) and \((\lambda_n)_{n\in\mathbb{N}}\) be a strictly increasing sequence of positive reals. It is assumed that the series \(\sum a_ne^{-\lambda_n z_0}\) converges.
Show that the series of functions \(\sum a_ne^{-\lambda_n z}\) converges uniformly on the set \(\mathcal{D}=\{z\in \mathbb{C},\quad \abs{\mathrm{arg}(z-z_0)}\leqslant \theta_0\}\cup\{z_0\}\), where \(\theta_0\in ]-\frac{\pi}{2},\frac{\pi}{2}[\).
Dirichlet series correspond to the unbounded \((\lambda_n)\) case.
Write \[ a_n e^{-\lambda_n z} = a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) } \]
Which theorem can be used to demonstrate uniform convergence when neither normal nor simple convergence is available?
We'll apply the uniform Cauchy criterion and look at the quantity: \[\sum_{n=N}^{N+p} a_n e^{-\lambda_n z}\]
Perform an Abel transformation by "integrating" \(a_n e^{-\lambda_n z_0}\) and "deriving" \(e^{-\lambda_n (z-z_0) }\).
We can pose \(z-z_0=a+i\mkern1mu b\), which makes it easy to translate membership of \(\mathcal{D}\setminus\{z_0\}\) by a simple majoration.
Find an upperbound for the modulus of the Cauchy packet as an expression independent of \(p,a,b\), which goes to 0 when \(N\to +\infty\).
Uniform Cauchy criterion
A sequence of functions of a set \(X\) in a normed ev \(E\) converges uniformly if and only if \[ \forall \epsilon>0,\exists n_0\in \mathbb{N},\forall p\geqslant n_0,\forall q\geqslant n_0,\forall x\in X,\quad \norm{f_p(x)-f_q(x)}\leqslant \epsilon \]
Uniform Cauchy criterion - Case of series
The criterion is directly reformulated as follows in the special case of function series:
\[ \forall \epsilon>0,\exists n_0\in \mathbb{N},\forall N\geqslant n_0,\forall p\in\mathbb{N},\forall x\in X,\quad \norm{\sum_{n=N}^{N+p} f_n(x)}\leqslant \epsilon \]
Note
The choice of \(n_0\) depends on \(\epsilon\), but not on \(p\) nor \(x\).
Note
The criterion is useful when we want to show uniform convergence in cases where simple convergence is not assumed, and so we have no candidate limit function.
In the case of series, it is used to show uniform convergence when there is no normal convergence.
To make the connection with the convergent series, it is natural to write: \[ a_n e^{-\lambda_n z} = a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) }\]
Then, to demonstrate uniform convergence without being able to assume simple convergence and without knowing the limit function, and without being able to go through normal convergence, we don't really have any tools other than the uniform Cauchy criterion; So we'll look at the expression, for \(z\in\mathcal{D}\): \[ \begin{alignat*}{1} \sum_{n=N}^{N+p} a_n e^{-\lambda_n z}&= \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) } \end{alignat*}\]
We can't find an upperbound for this sum directly, as we have no information about \(\sum \abs{a_n e^{-\lambda_n z_0}}\): it converges, but we can't assume absolute convergence of the series. On the other hand, to use the convergence hypothesis, we are forced to write \[ a_n e^{-\lambda_n z_0} = R_n - R_{n+1} \] where \[R_n = \sum_{k=n}^{+\infty} a_k e^{-\lambda_k z_0}\] and perform an Abel transform.
Let's write \(z-z_0=a+i\mkern1mu b\), which makes it easy to translate membership of \(\mathcal{D}\setminus\{z_0\}\) \[z\in \mathcal{D}\setminus\{z_0\} \Leftrightarrow a>0 \land \abs{\frac{b}{a}}\leqslant \abs{\tan{\theta_0}}\]
The Abel transformation is performed by "integrating" \(a_n e^{-\lambda_n z_0}\) and "deriving" \(e^{-\lambda_n (z-z_0) }\); \(\forall z\in \mathcal{D}\), \[ \begin{alignat*}{1} \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) } &= \sum_{n=N}^{N+p} (R_n-R_{n+1}) e^{-\lambda_n (z-z_0) } \\ &= \sum_{n=N}^{N+p} R_n e^{-\lambda_n (a+i\mkern1mu b) } - \sum_{n=N}^{N+p} R_{n+1} e^{-\lambda_n (a+i\mkern1mu b) } \\ &= \sum_{n=N}^{N+p} R_n e^{-\lambda_n (a+i\mkern1mu b) } - \sum_{n=N+1}^{N+p+1} R_{n} e^{-\lambda_{n-1} (a+i\mkern1mu b) } \\ &= R_N e^{-\lambda_N (a+i\mkern1mu b) } + \sum_{n=N+1}^{N+p} R_n (e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) } ) - R_{N+p+1} e^{-\lambda_{N+p} (a+i\mkern1mu b) }\\ \end{alignat*} \]
Now, it makes sense to take the modulus and majorize each term independently; the upperbound has to be an expression independent of \(p,a,b\), which tends to 0 when \(N\to +\infty\):\[ \begin{alignat*}{1} \abs{ \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) }} &\leqslant \abs{R_N} e^{-\lambda_N a} + \sum_{n=N+1}^{N+p} \abs{R_n}\abs{ e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) } } +\abs{ R_{N+p+1}} e^{-\lambda_{N+p} a } \\ \end{alignat*} \]
Let \(\epsilon>0\) and \(n_0\in\mathbb{N}\) be such that \(\forall n\geqslant n_0,\quad \abs{R_n}\leqslant \epsilon\).
Then, \(\forall N\geqslant n_0\),\[ \begin{alignat*}{1} \abs{ \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) }} &\leqslant \epsilon e^{-\lambda_N a} + \epsilon\sum_{n=N+1}^{N+p} \abs{ e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) } } +\epsilon e^{-\lambda_{N+p} a } \\ \end{alignat*} \]
Since the sequence \((\lambda_n)\) is positive and \(a>0\) we have: \[ \begin{alignat*}{1} \abs{ \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) }} &\leqslant 2\epsilon + \epsilon\sum_{n=N+1}^{N+p} \abs{ e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) } } \\ \end{alignat*} \]
To find an upperbound of the sum, compare to an integral; note that: \[ \begin{alignat*}{1} e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) }&= -(a+i\mkern1mu b)\int_{\lambda_{n-1}}^{\lambda_n}e^{-t(a+i\mkern1mu b)}\mathrm{d}t \end{alignat*} \]
And so, because \((\lambda_n)\) is increasing, \[ \begin{alignat*}{1} \abs{e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) }}&\leqslant \sqrt{a^2+b^2}\int_{\lambda_{n-1}}^{\lambda_n}\abs{e^{-t(a+i\mkern1mu b)}}\mathrm{d}t \\ &= \sqrt{a^2+b^2}\int_{\lambda_{n-1}}^{\lambda_n}e^{-ta}\mathrm{d}t \\ \end{alignat*} \] Note that the ordering of the integral bounds will be essential for applying the Chasles relation.
Then, \[ \begin{alignat*}{1} \sum_{n=N+1}^{N+p} \abs{ e^{-\lambda_n (a+i\mkern1mu b)}-e^{-\lambda_{n-1} (a+i\mkern1mu b) } } &\leqslant \sqrt{a^2+b^2} \sum_{n=N+1}^{N+p}\int_{\lambda_{n-1}}^{\lambda_n}e^{-ta}\mathrm{d}t \\ &= \sqrt{a^2+b^2} \int_{\lambda_{N}}^{\lambda_{N+p}}e^{-ta}\mathrm{d}t \\ &\leqslant \sqrt{a^2+b^2} \int_{\lambda_{N}}^{+\infty}e^{-ta}\mathrm{d}t \\ &= \sqrt{a^2+b^2} \frac{1}{a}e^{-\lambda_N a} \\ &= \sqrt{1+(\frac{b}{a})^2} e^{-\lambda_N a} \\ &\leqslant \sqrt{1+(\tan{\theta_0})^2} \\ &= \frac{1}{\cos{\theta_0}} \\ \end{alignat*} \] the latter not depending on \(a\), \(b\), \(p\).
We have proved, \(\forall N\geqslant n_0,\forall p\in\mathbb{N},\forall z\in\mathcal{D}\setminus\{z_0\}\), \[ \begin{alignat*}{1} \abs{ \sum_{n=N}^{N+p} a_n e^{-\lambda_n z_0 } e^{-\lambda_n (z-z_0) }} &\leqslant \epsilon(2+\frac{1}{\cos{\theta_0}}) \end{alignat*} \] majoration still valid when \(z=z_0\).
The series of functions \(\sum a_ne^{-\lambda_n z}\) converges uniformly on the set \(\mathcal{D}\).