5
$\begingroup$

Could you correct my following proof for the uniform convergence of the Weierstrass' M-Test?

At first, I'm going to write the preconditions:

We have a sequence of complex functions $(a_n(z))_{n \in \mathbb{N}_0}$ with $|a_n(z)|\leq M_n$ for $z \in K \subset \mathbb{C}$ and $\sum_{n=0}^\infty M_n < \infty$.

This implies that $L(z)= \sum_{n=0}^{\infty} a_n(z)$ is uniformly convergent (which means that for all $z \in K$ there is an $\epsilon >0$ and $m_0>0$ such that $|L(z)-\sum_{n=0}^{m} a_n(z)|<\epsilon$ , $m\geq m_0$, $z \in K$.)

I have to show this following two steps:

1) The series is uniformly cauchy, i.e. there is an $\epsilon >0$ and $m_0$ such that $|\sum_{n=0}^{M_2} a_n(z) - \sum_{n=0}^{M_1} a_j(z)|<\epsilon$, $M_2>M_1\geq m_0$, $z \in K$.

2) Given the existence of the Limit of the partial sums, show that we get the uniform convergence of the partial sums.

Now I am going to present my ideas:

For step 1): I already proved that the series $\sum_{n=0}^{\infty} a_n(z)$ converges absolutely. And therefore, it converges. My idea is: Let $s(z)$ be the limit of the series. Then for all $\epsilon>0$ there is an $m_0 \in \mathbb{N}$ such that $|\sum_{n=0}^{M_1}a_n(z)-s(z)|<\frac{\epsilon}{2}$ and $|\sum_{n=0}^{M_2}a_n(z)-s(z)|<\frac{\epsilon}{2}$ for all $M_2,M_1\geq m_0$. And then it follows that $|\sum_{n=0}^{M_2} a_n(z) - \sum_{n=0}^{M_1} a_j(z)|=|\sum_{n=0}^{M_1}a_n(z)-s(z)+s(z)-\sum_{n=0}^{M_2}a_n(z)|\leq |\sum_{n=0}^{M_1}a_n(z)-s(z)|+|\sum_{n=0}^{M_2}a_n(z)-s(z)|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$. Is this correct?

For step 2): By convergence of the series we know that for all $\epsilon>0$ there is an $m_0 \in \mathbb{N}$ such that $|\sum_{n=0}^{M_1}a_n(z)-s(z)|<\frac{\epsilon}{2}$ for all $M_1>m_0$. By step 1) we know that for all $\epsilon$ is an $m_0$ gibt such that $|\sum_{n=0}^{M_2} a_n(z) - \sum_{n=0}^{M_1} a_j(z)|<\epsilon$ for all $M_2>M_1\geq m_0$. Hence, we get: $|s(z)-\sum_{n=0}^{M_1}a_n(z)|\leq |s(z)-\sum_{n=0}^{M_2}a_n(z)|+|\sum_{n=0}^{M_2}a_n(z)-\sum_{n=0}^{M_1}a_n(z)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Is this correct? If it's not correct, how does the proof work?

Thank you!!

$\endgroup$
  • $\begingroup$ Your $m_0$ in 1) may depend on $z$ so your proof is not valid. $\endgroup$ – Kavi Rama Murthy Nov 18 '19 at 8:50
4
$\begingroup$

You have not used the Weierstrass conditions for anything other than establishing pointwise absolute convergence. Hence, your subsequent steps to prove uniform convergence fall short.

As you began in step (1), for each $z \in K$ we have $a_n(z) \leqslant M_n$ and $\sum_{n \geqslant 1} M_n < \infty$ and it follows by the comparison test that there exists $S(z) = \sum_{n=1}^\infty a_n(z)$ for each $z \in K$.

Now we apply the Weierstrass condition to show that the series is uniformly Cauchy. Since $\sum_{n \geqslant 1} M_n$ is convergent and the partial sums form a Cauchy sequence, for any $\epsilon > 0$ there exists $N(\epsilon)$ (independent of $z$) such that for all $m > n > N(\epsilon)$ and all $z \in K$ we have

$$\tag{1}|S_m(z) - S_n(z)| = \left|\sum_{k=n+1}^m a_k(z) \right| \leqslant\sum_{k=n+1}^m |a_k(z)| \leqslant \sum_{k = n+1}^m M_n < \epsilon$$

Completing the proof that $S_n(z) \to S(z)$ uniformly on $K$ is a as simple as asserting that for all $n > N(\epsilon)$ and for all $z \in K$ we have

$$\tag{2}|S(z) - S_n(z)| = \lim_{m\to \infty}|S_m(z) - S_n(z)| \leqslant \epsilon$$

To provide justification for (2), by the reverse triangle inequality,

$$\left|\, |S_m(z) - S_n(z)| - |S(z) - S_n(z)| \,\right| \leqslant |S_m(z) - S(z)|, $$

and it follows that the pointwise convergence $S_m(z) \to S(z)$ implies that

$$\tag{3} \lim_{m \to \infty}|S_m(z) - S_n(z)| = |S(z) - S_n(z)|$$

Suppose that for some $z \in K$ and $n > N(\epsilon)$ we have $|S(z) - S_n(z)| = \alpha > \epsilon$. By the convergence expressed in (3) there exists $M \in \mathbb{N}$ such that for all $m > M$ we have

$$\left|\,|S_m(z) - S_n(z)| - |S(z) - S_n(z)|\, \right| < \frac{\alpha - \epsilon}{2},$$

which implies

$$|S_m(z)- S(z)| > |S(z) - S_n(z)| - \frac{\alpha- \epsilon}{2} = \alpha - \frac{\alpha- \epsilon}{2} = \frac{\alpha+ \epsilon}{2} > \epsilon $$

This contradicts (1) when $m > \max(M, N(\epsilon))$, and, thus, (2) must hold and the convergence of the series to $S(z)$ is uniform.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.