3. 함수열과 함수급수Sequence and Series of Functions

The claim is false. We give a counterexample. Let $(a, b) := (-1, 1)$ and for all $n \in \mathbb{N}$, define

Since $1 + nx^2 \geq 1 > 0$, $f_n$ is differentiable on $(-1, 1)$ (indeed on all of $\mathbb{R}$).

First we find the pointwise limit. If $x = 0$, then $f_n(0) = 0 \to 0$. If $x \neq 0$, then as $n \to \infty$, $1 + nx^2 \to \infty$, so

Thus $\lim_{n \to \infty} f_n(x) = 0$ exists for all $x \in (-1, 1)$, and

The function $f \equiv 0$ is differentiable on $(-1, 1)$ with $f'(x) = 0$, in particular $f'(0) = 0$.

Next we compute $f_n'$.

If $x = 0$, then $f_n'(0) = \dfrac{1 - 0}{(1+0)^2} = 1$ for all $n$, so

If $x \neq 0$, then as $n \to \infty$, $nx^2 \to \infty$; letting $m := nx^2$,

that is, $\lim_{n \to \infty} f_n'(x) = 0$. Hence $\lim_{n \to \infty} f_n'(x)$ exists for all $x \in (-1, 1)$.

Thus $f_n$ satisfies all the hypotheses of this exercise: $\lim_{n\to\infty} f_n(x)$ and $\lim_{n\to\infty} f_n'(x)$ exist for all $x \in (-1,1)$, and $f = \lim_{n\to\infty} f_n$ is differentiable on $(-1,1)$. However,

so the claimed equality fails. This disproves the claim.

(Note that $f_n(x) = \sin(nx)/\sqrt{n}$ from Example 3.3 cannot serve as this counterexample. Since $\lim_{n \to \infty} f_n'(0) = \lim_{n \to \infty} \sqrt{n} = \infty$ does not exist as a real number, Example 3.3 fails to satisfy the very hypothesis of this exercise that $\lim_{n\to\infty} f_n'(x)$ exist for all $x$. This is exactly why a new counterexample, in which $\lim_{n\to\infty} f_n'$ exists everywhere yet still $f' \neq \lim_{n\to\infty} f_n'$, is needed above.)

• (i) Suppose $f_n$ converges to $f$ uniformly on $E$. Fix $x \in E$. Let $\varepsilon > 0$. By the definition of uniform convergence (Definition 3.7), there exists $N \in \mathbb{N}$ such that for all $n \geq N$,
  $$\sup_{y \in E} |f_n(y) - f(y)| \lt \varepsilon.$$

  Since $x \in E$, for all $n \geq N$,

  $$|f_n(x) - f(x)| \leq \sup_{y \in E} |f_n(y) - f(y)| \lt \varepsilon.$$

  That is, $f_n(x) \to f(x)$. Since $x \in E$ was arbitrary, $f_n$ converges to $f$ pointwise on $E$.

• (ii) We show the converse fails, by a counterexample. Let $E := [0, 1]$ and for all $n \in \mathbb{N}$, let $f_n(x) := x^n$.

  We find the pointwise limit. If $x \in [0, 1)$, then $0 \leq x \lt 1$, so $x^n \to 0$. If $x = 1$, then $x^n = 1 \to 1$. Thus

  $$f(x) := \lim_{n \to \infty} f_n(x) = \begin{cases} 0 & (0 \leq x \lt 1) \\ 1 & (x = 1) \end{cases}$$

  and $f_n$ converges to $f$ pointwise on $[0, 1]$.

  We now show that this convergence is not uniform, using the criterion of Theorem 3.9. Let $\varepsilon := 1/2$. For any $N \in \mathbb{N}$, let

  $$x_N := \left(\frac{1}{2}\right)^{1/N} \in (0, 1).$$

  Then $x_N \in [0, 1)$, so $f(x_N) = 0$, and

  $$|f_N(x_N) - f(x_N)| = |x_N^N - 0| = \left(\left(\frac{1}{2}\right)^{1/N}\right)^N = \frac{1}{2}.$$

  That is, $|f_N(x_N) - f(x_N)| = 1/2$, which is not less than $\varepsilon = 1/2$. Hence no matter which $N$ we choose, it is impossible that $|f_n(x) - f(x)| \lt \varepsilon$ for all $n \geq N$ and $x \in [0,1]$ (taking $n = N$ and $x = x_N$ gives a counterexample). By Theorem 3.9, $f_n$ does not converge to $f$ uniformly on $[0,1]$. This disproves the converse.