Exercise 8.2.17

Question

\begin{enumerate}[label=(\alph*)] 
\item The sequence $(x_k)$ does not necessarily converge, but explain why there exists a subsequence $(x_{k_l})$ that is convergent. Let $x = \lim(x_{k_l})$.
\item Prove that $f_{k_l}(x_{k_l}) \to f(x)$.
\item Now finish the proof that $A_{m,n}$ is closed.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item The sequence $(x_k)$ is in $\mathbf{R}$ and is bounded, so by the Bolzano-Weierstrauss Theorem it contains a convergent subsequence. \item Let $\epsilon > 0$ be arbitrary. $f$ is continuous and $(x_{k_l}) \to x$, so for some $K$, for $k_l > K_1$, $\left|f(x_{k_l}) - f(x)\right| < \epsilon / 2$. We also have $(f_k)$ uniformly converging to $f$, so for some $K_2$ and $k_l > K_2$, $\left|f(x_{k_l}) - f_{k_l}(x_{k_l})\right| < \epsilon/2$. Thus if $k_l > \max\{K_1,K_2\}$ then $\left|f(x) - f_{k_l}(x_{k_l})\right| < \epsilon$. \item For conciseness, I'll abuse notation a bit and redefine $(x_k)$ to be the convergent subsequence $(x_{k_l})$ in part (a) and (b). Let $y \in V_{1/m}(x)$. Let $\epsilon > 0$ be arbitrary. Choose a $x_k$ ``close'' to $x$ (we'll specify exactly how close in a moment). We have $$\left|f(y) - f(x)\right| \leq \left|f(x) - f(x_k)\right| + \left|f(x_k) - f_k(x_k)\right| + \left|f_k(x_k) - f_k(y)\right| + \left|f_k(y) - f(y)\right|$$ The first term on the right can be made small by continuity on $f$; for an arbitrary $\epsilon_1$ we can choose $x_k \in V_{\delta_1}(x)$ to ensure $|f(x) - f(x_k)| < \epsilon_1$. The second and fourth terms can be made small because of uniform convergence of $(f_k)$; we can identify $\delta_2$ so that $x_k \in V_{\delta_2}(x)$ implies $\left|f_k - f\right| < \epsilon_2$ for arbitrary $\epsilon_2 > 0$. For the third term, first note that if $\delta_3 < 1/m - \left|y-x\right|$, then $y \in V_{\delta_3}(x_k)$. Now since $f_k \in A_{m,n}$, $$\left|f_k(x_k) -f_k(y)\right| \leq n \left|y - x_k\right| \leq n \left|y-x\right| + n \left|x - x_k\right|$$ Putting it all together and rearranging, we have $$\left|\frac{f(y) - f(x)}{y-x}\right| \leq n + \frac{\epsilon_1 + 2\epsilon_2 + n \left|x - x_k\right|}{\left|y-x\right|} $$ If we set $\epsilon_1 = \epsilon_2 = \epsilon / |y-x|$, define $$\delta_4 = \frac{\left|y-x\right|}{n}$$ and $\delta = \min\{\delta_1, \delta_2, \delta_3, \delta_4 \}$, and we find $x_k$ so that $x_k \in V_\delta(x)$, then we'll have $$\left|\frac{f(y) - f(x)}{y-x}\right| \leq n + \epsilon$$ for arbitrary $\epsilon > 0$, implying $$\left|\frac{f(y) - f(x)}{y-x}\right| \leq n $$ and that $f \in A_{m,n}$ and $A_{m,n}$ is closed. \end{enumerate}

Exercise 8.2.17

Answers

Comments

Exercise 8.2.17

Answers

Comments

Add answer