Problem 1

Exercise 2.1 (Fundamental Limit Interchange)   Suppose $f_n\to f$ uniformly and $\lim_{x\to x_0}f_n(x)$ exists for each $f_n$. Then

$$\lim_{x\to x_0}\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} \lim_{x\to x_0} f_n(x)$$ (2.1)

Proof. Let $L_n := \lim_{x\to x_0} f_n(x)$. Then we are going to show

$$\lim_{x\to x_0} \lim_{n\to \infty} f_n(x) = \lim_{n\to\infty} L_n$$ (2.2)

Equivalently

$$\lim_{x\to x_0} \lim_{n\to \infty} \vert f_n(x)-L_n\vert = 0$$ (2.3)

Select $N$ so that $\Vert f_n-f_m\Vert<\epsilon/3$ for all $n,m\geq N$. Select $\delta>0$ so that $\vert f_N(x)-L_N\vert<\epsilon/3$ for all $0<\vert x-x_0\vert<\delta$. For any $n>N$, another distance can be estimated:

$$\vert f_n(x) - L_n\vert = \vert f_n(x) - f_N(x) + f_N(x) - L_N + L_N - L_n\vert$$ (2.4)

$$\leq \vert f_n(x) - f_N(x)\vert + \vert f_N(x) - L_N\vert + \vert L_N - L_n\vert$$ (2.5)

The first summand and the second summand are bounded by convergence and continuity, respectively. The final summand is bounded by Cauchiness. Verify:

$$\vert L_N - L_n\vert = \lim_{x\to x_0} \vert f_N(x)-f_n(x)\vert \leq \Vert f_N-f_n\Vert$$ (2.6)

Therefore

$$\lim_{x\to x_0}\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} \lim_{x\to x_0} f_n(x)$$ (2.7)

$\qedsymbol$