Exercise 6.2.6

Question

Assume $f_{n} \rightarrow f$ on a set $A$. Theorem 6.2.6 is an example of a typical type of question which asks whether a trait possessed by each $f_{n}$ is inherited by the limit function. Provide an example to show that all of the following propositions are false if the convergence is only assumed to be pointwise on $A$. Then go back and decide which are true under the stronger hypothesis of uniform convergence.
  \begin{enumerate}[label=(\alph*)] 
  \item If each $f_{n}$ is uniformly continuous, then $f$ is uniformly continuous.
  \item If each $f_{n}$ is bounded, then $f$ is bounded.
  \item If each $f_{n}$ has a finite number of discontinuities, then $f$ has a finite number of discontinuities.
  \item If each $f_{n}$ has fewer than $M$ discontinuities (where $M \in \mathbf{N}$ is fixed), then $f$ has fewer than $M$ discontinuities.
  \item If each $f_{n}$ has at most a countable number of discontinuities, then $f$ has at most a countable number of discontinuities.

\end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item False pointwise when $$ f_n(x) = \frac{1}{1+nx^2} \quad\text{and}\quad f(x) = \begin{cases} 1 &\text{if $x = 0$} \ 0 &\text{otherwise} \end{cases} $$ Now suppose $(f_n)$ converges to $f$ uniformly. We have $$|f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| $$ By uniform convergence to $f$, we can fix $n$ large enough so that the first and last terms are both less than $\epsilon / 3$. Then since $f_n$ is uniformly continuous, we can choose $\delta$ so that $|x - y| < \delta$ implies the middle term is less than $\epsilon / 3$, and that $|f(x) - f(y)| < \epsilon$, implying $f$ is uniformly continuous. \item False pointwise when $$ f_n(x) = \begin{cases} x &\text{if $x < n$} \ 0 &\text{otherwise} \end{cases} \quad\text{and}\quad f(x) = x $$ Now suppose $(f_n) \to f$ uniformly. We want to show $f$ is bounded. Let $M_n$ bound $f_n$, ie. $|f_n(x)| N$ implies $$ |f_n(x) - f_m(x)| < 1 $$ Setting $n=N$ and rearranging gives $$ |f_m(x)| < |f_N(x)| + \epsilon < M_N + \epsilon $$ implying $f_m$ is bounded by $M_N$ when $m \ge N$. Now set $\tilde N > N$ large enough that $m \ge N$ implies $|f(x)-f_m(x)|<1$ which, after rearranging gives $$ |f(x)| < 1 + |f_m(x)| < 1 + M_N $$ Implying $f$ is bounded. \item False for pointwise and uniform convergence. Define $$ f_n(x) = \begin{cases} 1/m &\text{if $x = 1/m$ and $m\le n$, $m \in\mathbf{N}$} \ 0 &\text{otherwise} \end{cases} $$ and $$ f(x) = \begin{cases} 1/m &\text{if $x=1/m$, $m \in \mathbf{N}$} \ 0 &\text{otherwise} \end{cases} $$ The reason $(f_n) \to f$ uniformly is $$ |f(x) - f_n(x)| = \left( \begin{cases} 1/m &\text{if $x=1/m$ and $m > n$, $m \in \mathbf{N}$} \ 0 &\text{otherwise} \end{cases} \right) < 1/{(n+1)} $$ Each $f_n$ has exactly $n$ point discontinuities, but $f$ has countably many. Thus this is a counterexample. Intuitively $f_n$ adds on finer and finer details of $f$, which is why it converges uniformly. But discontinuities can be as small/detailed as we want without screwing up uniform convergence. \item False pointwise since we can have each $(f_n)$ continuous (zero discontinuities) but have $f$ not be continuous (see (a) for an example). Now suppose $(f_n) \to f$ uniformly. Let $x_0$ be a discontinuity of $f$, meaning there exists an $\epsilon_0$ such that $|f(x_0)-f(x)| > \epsilon_0$ no matter how small $|x-x_0|$ is. I'd like to show $x_0$ is a discontinuity of $f_n$ for some $n$, i.e.\ that there exists an $\epsilon_0'>0$ such that $|f_n(x_0)-f_n(x)|>\epsilon_0'$ no matter how small $|x-x_0|$ is. Pick $\epsilon<\epsilon_0/2$ and set $N$ large enough that $n \ge N$ implies $|f_n(x)-f(x)|<\epsilon$. Applying the three way triangle inequality gives $$ \begin{aligned} \epsilon_0 &< |f(x_0) - f(x)| \ &\le |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f_n(x) - f(x)| \ &< 2\epsilon + |f_n(x_0) - f_n(x)| \end{aligned} $$ Letting $\epsilon_0' = \epsilon_0-2\epsilon > 0$ we see $|f_n(x_0) - f_n(x)| > \epsilon_0'$ meaning $f_n$ is not continuous at $x_0$ for all $n \ge N$. Now given discontinuities of $f$, applying the above process multiple times shows that \emph{eventually} $f_n$ will have every discontinuity of $f$, but $f_n$ has at most $M$ discontinuities, implying $f$ has at most $M$ discontinuities. \item False pointwise when (using a modified version of Thomae's function for $f_n$) $$ f_n(x) = \begin{cases} 1 &\text{if $x = 0$} \ \frac{n}{n+q} &\text{if $x = p/q$ (in lowest terms)} \ 0 &\text{otherwise} \end{cases} \quad\text{and}\quad f(x) = \begin{cases} 1 &\text{if $x \in \mathbf{Q}$} \ 0 &\text{if $x \notin \mathbf{Q}$} \ \end{cases} $$ Now suppose $(f_n) \to f$ uniformly. In (d) we showed every discontinuity of $f$ is eventually a discontinuity of $f_n$; reworded this is saying (where $D_f$ is the set of discontinuities of $f$) $$ D_f \subseteq \bigcup_{n=1}^\infty D_{f_{n}} $$ Since each $D_{f_n}$ is countable, this implies $D_f$ is countable. \end{enumerate}

Exercise 6.2.6

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Exercise 6.2.6

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