Exercise 6.2.9

Question

Assume $\left(f_{n}\right)$ and $\left(g_{n}\right)$ are uniformly convergent sequences of functions.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that $\left(f_{n}+g_{n}\right)$ is a uniformly convergent sequence of functions.
  \item Give an example to show that the product $\left(f_{n} g_{n}\right)$ may not converge uniformly.
  \item Prove that if there exists an $M>0$ such that $\left|f_{n}\right| \leq M$ and $\left|g_{n}\right| \leq M$ for all $n \in \mathbf{N}$, then $\left(f_{n} g_{n}\right)$ does converge uniformly.

\end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Obvious by the triangle inequality and Cauchy Criterion
  \item Let $f_n(x) = x = f(x)$ and $g_n(x) = x+1/n$. Suppose $n,m \ge N$ for some $N$, Cauchy gives us
    $$
    |f_ng_n - f_mg_m| = |x(1/n - 1/m)|
    $$
    Making $x$ large makes the error blow up regardless of how big $N$ is, thus $f_ng_n$ does not converge uniformly.
  \item By the triangle inequality (same trick as for the product rule)
    $$
    \begin{aligned}
      |f_ng_n - f_mg_m|
      &\le |f_ng_n - f_ng_m| + |f_ng_m - f_mg_m| \
      &= |f_n|\cdot|g_n - g_m| + |g_m|\cdot |f_n-f_m| \
      &< M|g_n - g_m| + M|f_n-f_m| \
      &< \epsilon/2 + \epsilon/2 = \epsilon
    \end{aligned}
    $$
    After setting $N$ big enough that $n,m \ge N$ implies $|f_n-f_m|<M/2\epsilon$ and $|g_n-g_m|<M/2\epsilon$.
   \end{enumerate}

Exercise 6.2.9

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Exercise 6.2.9

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