If \(f\) is continuous, is it necessarily the case that \(f'\) is continuous?
If \(f_n \to f\), is it necessarily the case that \(f_n'\) converges to \(f'\) (or at all)?
Is it true that the sum of differentiable functions is differentiable?
Is it true that the limit of integrals equals the integral of the limit?
Is it true that a limit of continuous functions is continuous?
Show that a subset of a metric space is closed iff it is complete.
Show that if \(m(E) < \infty\) and \(f_n\to f\) uniformly, then \(\lim \int_E f_n = \int_E f\).
The closed-iff-complete statement needs the ambient space to be complete: \((0,1]\) is closed in the metric space \((0,1]\) but is not complete, since \(1/n\) is Cauchy with no limit in it. Complete subsets are closed in every metric space.
Solution
1. \(f(x) = x^2\sin(1/x)\), \(f(0) = 0\), is differentiable on \(\RR\) and \(f'\) is discontinuous at \(0\).
\(\frac{f(h) - f(0)}{h} = h\sin(1/h) \to 0\), so \(f'(0) = 0\). For \(x \ne 0\), \(f'(x) = 2x\sin(1/x) - \cos(1/x)\), and \(f'(1/(2k\pi)) = -1\) for every \(k \geq 1\).
2. \(f_n(x) = \frac{\sin(nx)}{n} \to 0\) uniformly, while \(f_n'(x) = \cos(nx)\) satisfies \(f_n'(0) = 1\) for all \(n\) and \(f_n'(\pi) = (-1)^n\).
\(|f_n| \le 1/n\). So \(f_n'(0) \to 1 \neq 0\), the derivative of the limit, and \(f_n'(\pi)\) does not converge.
3. If \(f\) and \(g\) are differentiable at \(x\), so is \(f + g\), with \((f + g)'(x) = f'(x) + g'(x)\).
The difference quotient of \(f + g\) is the sum of the difference quotients of \(f\) and \(g\).
4. \(f_n = n\chi_{(0, 1/n)}\) on \([0,1]\) satisfies \(f_n \to 0\) pointwise and \(\int f_n = 1\) for all \(n\).
For each \(x \in [0,1]\), \(f_n(x) = 0\) once \(1/n \le x\) or if \(x = 0\), and \(\int f_n = n \cdot \frac1n\).
5. \(f_n(x) = x^n\) on \([0,1]\) are continuous and converge pointwise to \(\chi_{\theset{1}}\), which is discontinuous at \(1\).
\(x^n \to 0\) for \(0 \le x < 1\) and \(1^n = 1\). By the uniform limit theorem the convergence is not uniform.
6. Let \(A\) be a subset of a metric space \(X\). If \(A\) is complete, then \(A\) is closed. If \(X\) is complete and \(A\) is closed, then \(A\) is complete.
If \(A\) is complete and \(x_k \in A\) with \(x_k \to x \in X\), then \((x_k)\) is Cauchy, so it converges in \(A\), and by uniqueness of limits \(x \in A\). If \(X\) is complete and \(A\) is closed, a Cauchy sequence in \(A\) converges to some \(x \in X\), and \(x \in A\) because \(A\) is closed.
7. If \(m(E) < \infty\) and \(f_n \to f\) uniformly on \(E\), then \(\lim_n \int_E f_n = \int_E f\).
\(\left|\int_E f_n - \int_E f\right| \le \int_E |f_n - f| \le m(E)\sup_E|f_n - f| \to 0\).