Exercise 6.4.2

Question

Decide whether each proposition is true or false, providing a short justification or counterexample as appropriate.
  \begin{enumerate}[label=(\alph*)] 
  \item If $\sum_{n=1}^{\infty} g_{n}$ converges uniformly, then $\left(g_{n}\right)$ converges uniformly to zero.
  \item If $0 \leq f_{n}(x) \leq g_{n}(x)$ and $\sum_{n=1}^{\infty} g_{n}$ converges uniformly, then $\sum_{n=1}^{\infty} f_{n}$ converges uniformly.
  \item If $\sum_{n=1}^{\infty} f_{n}$ converges uniformly on $A$, then there exist constants $M_{n}$ such that $\left|f_{n}(x)\right| \leq M_{n}$ for all $x \in A$ and $\sum_{n=1}^{\infty} M_{n}$ converges.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item True: applying the Cauchy Criterion with $n = m+1$ we have that $|g_n(x)| < \epsilon$ for any $\epsilon > 0$, therefore $(g_n) \to 0$.
    \item True: $$\left|\sum^n_{i=m+1}f_i(x)\right| = \sum^n_{i=m+1}f_i(x) \leq \sum^n_{i=m+1}g_i(x) = \left|\sum^n_{i=m+1} g_i(x)\right| < \epsilon$$ and therefore $\sum(f_i)$ converges uniformly.
    \item False: Consider the following sequence of functions, defined over $[0,1)$:
    $$g_{i,j}(x) = \begin{cases}
        2^{-i} & 2^{-i}(j-1) \leq x < 2^{-i}j \
        0 & \text{otherwise}
    \end{cases}$$
    with $i \geq 1$ and $j$ an integer ranging from 1 to $2^i$ inclusive. Each $g_{i,j}(x)$ consists of a pulse of height and width $2^{-i}$, at disjoint locations for each $i$. Let $f_n(x)$ be obtained by iterating through each $g_{1,j}$, then through each $g_{2,j}$, then through each $g_{3,j}$, and so on.

$\sum^\infty_{n=1} f_n$ converges to 1 because
   $$\sum^{2^i}_{k=1} g_{i,k} = 2^{-i}$$
   , and uniform convergence is achieved when we include all of the $g_{i,j}$ for a given $g_i$. On the other hand, the upper bound (and therefore minimum value of the constant $M_{n}$) for each $g_{i,j}$ is $2^{-i}$, with
   $$\sum^{2^i}_{k=1} \max g_{i,k}(x) = 1$$
   which implies that $\sum^\infty_{n=1} M_n$ will not converge.
   \end{enumerate}

Exercise 6.4.2

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Exercise 6.4.2

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