Find a function that is differentiable but not continuously differentiable.
Prove the uniform limit theorem: a uniform limit of continuous function is continuous.
Show that the uniform limit of bounded functions is uniformly bounded.
Construct sequences of functions \(\ts{f_n}_{n\in \NN}\) and \(\ts{g_n}_{n\in \NN}\) which converge uniformly on some set \(E\), and yet their product sequence \(\ts{h_n}_{n\in \NN}\) with \(h_n \da f_n g_n\) does not converge uniformly.
- Show that if \(f_n, g_n\) are additionally bounded, then \(h_n\) does converge uniformly.
Find a sequence of functions such that \[\frac{d}{d x} \lim _{n \rightarrow \infty} f_{n}(x) \neq \lim _{n \rightarrow \infty} \frac{d}{d x} f_{n}(x)\]
Find a uniform limit of differentiable functions that is not differentiable.
Prove that the Cantor set is a Borel set.
Show the Cantor ternary set is totally disconnected; that is show it contains no nonempty open interval.
Solution
1. \(f(x) = x^2\sin(1/x)\) for \(x \ne 0\), \(f(0) = 0\), is differentiable on \(\RR\) but not continuously differentiable.
1.1. \(f\) is differentiable everywhere, with \(f'(0) = 0\) and \(f'(x) = 2x\sin(1/x) - \cos(1/x)\) for \(x \neq 0\).
\(\frac{f(h) - f(0)}{h} = h\sin(1/h)\) and \(|h\sin(1/h)| \leq |h| \to 0\). For \(x \neq 0\), \(f\) is a product and composition of differentiable functions near \(x\), and the product and chain rules give the formula.
1.2. \(f'\) is discontinuous at \(0\).
At \(x_k = 1/(2k\pi)\), \(f'(x_k) = -1\) for every \(k\), while \(x_k \to 0\) and \(f'(0) = 0\).
1.3. Q.E.D.
Steps 1 and 2.
2. If \(f_n \to f\) uniformly on \(E\) and each \(f_n\) is continuous, then \(f\) is continuous.
2.1. Fix \(x_0 \in E\) and \(\eps > 0\). There are \(n\) with \(\|f - f_n\|_\infty < \eps/3\) and \(\delta > 0\) with \(|f_n(x) - f_n(x_0)| < \eps/3\) for \(x \in E\), \(|x - x_0| < \delta\).
The first is uniform convergence; the second is continuity of \(f_n\) at \(x_0\).
2.2. Q.E.D.
For \(x \in E\) with \(|x - x_0| < \delta\), step 1 and the triangle inequality give \(|f(x) - f(x_0)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \eps\).
3. If \(f_n \to f\) uniformly and each \(f_n\) is bounded, then \(f\) is bounded, and \(\sup_{n}\|f_n\|_\infty < \infty\).
Choose \(N\) with \(\|f - f_n\|_\infty \le 1\) for \(n \geq N\). Then \(\|f\|_\infty \le \|f_N\|_\infty + 1\), and \(\|f_n\|_\infty \leq \|f\|_\infty + 1\) for \(n \geq N\), so \(\sup_n \|f_n\|_\infty \leq \max(\|f_1\|_\infty, \ldots, \|f_{N-1}\|_\infty, \|f\|_\infty + 1)\).
4. Products of uniformly convergent sequences need not converge uniformly; they do if every \(f_n\) and \(g_n\) is bounded.
4.1. On \(E = [1, \infty)\), \(f_n(x) = x + \frac{1}{n}\) and \(g_n(x) = x\) converge uniformly, but \(h_n = f_n g_n\) does not.
\(\sup_{x\ge1}|f_n(x) - x| = 1/n \to 0\), and \(g_n\) is constant in \(n\). But \(h_n(x) = x^2 + \frac{x}{n} \to x^2\) pointwise, and \(\sup_{x\ge1}|h_n(x) - x^2| = \sup_{x\ge1}\frac{x}{n} = \infty\) for every \(n\).
4.2. If \(f_n \to f\) and \(g_n \to g\) uniformly and every \(f_n\) and \(g_n\) is bounded, then \(f_n g_n \to fg\) uniformly.
By step 3, \(g\) is bounded and there is \(M\) with \(\|f_n\|_\infty \le M\) for all \(n\) and \(\|g\|_\infty \le M\). Then \(|f_n g_n - fg| \le |f_n|\,|g_n - g| + |g|\,|f_n - f| \le M\|g_n - g\|_\infty + M\|f_n - f\|_\infty \to 0\) uniformly on \(E\).
4.3. Q.E.D.
Steps 1 and 2.
5. \(f_n(x) = \frac{\sin(nx)}{n}\) satisfies \(\frac{d}{dx}\lim_n f_n \ne \lim_n \frac{d}{dx}f_n\) at \(x = 0\).
\(|f_n| \leq 1/n\), so \(f_n \to 0\) uniformly and \(\frac{d}{dx}\lim_n f_n(x) = 0\). But \(f_n'(x) = \cos(nx)\), so \(\lim_n f_n'(0) = 1 \ne 0\).
6. \(f_n(x) = \sqrt{x^2 + \frac{1}{n}}\) are differentiable on \(\RR\) and converge uniformly to \(|x|\), which is not differentiable at \(0\).
6.1. \(f_n \to |x|\) uniformly.
\(0 \le \sqrt{x^2 + 1/n} - |x| = \frac{1/n}{\sqrt{x^2 + 1/n} + |x|} \le \frac{1/n}{\sqrt{1/n}} = \frac{1}{\sqrt n} \to 0\).
6.2. Each \(f_n\) is smooth, and \(|x|\) is not differentiable at \(0\).
\(x \mapsto \sqrt{x^2 + c}\) is smooth for \(c > 0\), since \(x^2 + c > 0\). The function \(|x|\) has left derivative \(-1\) and right derivative \(1\) at \(0\).
6.3. Q.E.D.
Steps 1 and 2.
7. The Cantor set \(C\) is a Borel set.
\(C = \bigcap_n C_n\), where \(C_n\) is the union of the \(2^n\) closed intervals of length \(3^{-n}\) remaining after \(n\) stages. Each \(C_n\) is closed, so \(C\) is closed, and closed sets are Borel.
8. \(C\) contains no nonempty open interval.
\(m(C) \le m(C_n) = (2/3)^n\) for all \(n\), so \(m(C) = 0\). A nonempty open interval has positive measure, so it is not contained in \(C\).