Created on 2023/04/07

(Refer to the Textbook:Bartle & sherbert, Introduction to Real Analysis, 4th Ed, Wiley)


Let $\left(f_n\right),\left(g_n\right)$ be sequences of bounded functions on $A$ that converge uniformly on $A$ to $f, g$, respectively. Show that $\left(f_n g_n\right)$ converges uniformly on $A$ to $f g$.


It is easy to show that the uniform norm $||\cdot||_A$ satisfies triangle inequality. Therefore we have

$||f_ng_n-fg||_A\leq ||f_ng_n-f_ng||_A+||f_n g -fg||_A\leq ||f_n||_A||g_n-g||_A+||g||_A||f_n-f||_A$.

By definition of uniform convergence, for $\epsilon>0, \exists N_\epsilon, \forall n>N_\epsilon,||f_n-f||_A<\epsilon ,||g_n-g||_A<\epsilon$.

Suppose $n>N_\epsilon$, then

$||f||_A\leq ||f-f_n||_A+||f_n||_A<\epsilon+||f_n||_A$, so $f$ is bounded. Similarly we know $g$ is bounded.

$||f_n||_A\leq ||f_n-f||_A+||f||_A<\epsilon+||f||_A$, so for $n$ large enough, $f_n$ are uniformly bounded.

Therefore, $||f_ng_n-fg||_A\leq (\epsilon+||f||_A)\epsilon+||g||_A\epsilon$. For any $\epsilon'>0$, we may pick $\epsilon<\mathrm{min}\{\frac{\epsilon'}{1+||f||_A+||g||_A},1\}$. Then for $n>N_\epsilon$, we have $||f_ng_n-fg||_A\leq \epsilon'$. Q.E.D


Let $\left(f_n\right)$ be a sequence of functions that converges uniformly to $f$ on $A$ and that satisfies $\left|f_n(x)\right| \leq$ $M$ for all $n \in \mathbb{N}$ and all $x \in A$. If $g$ is continuous on the interval $[-M, M]$, show that the sequence $\left(g \circ f_n\right)$ converges uniformly to $g \circ f$ on $A$.


For any $\epsilon>0$, there exists $\delta_\epsilon>0$, if $x,y\in[-M,M]$ and $|x-y|\leq \delta_\epsilon$, then $|g(x)-g(y)|<\epsilon$. Let $N_\delta$ be the integer such that $\forall n>N_\delta$, $||f_n-f||_A\leq \delta_\epsilon$. Since $|f_n(x)-f(x)|\leq ||f_n-f||_A\leq \delta_\epsilon$, then $||g \circ f_n-g\circ f||_A=\operatorname{sup}_{x\in A}|g \circ f_n(x)-g\circ f(x)|<\epsilon$. Q.E.D.


Construct a sequence of functions on $[0,1]$ each of which is discontinuous at every point of $[0,1]$ and which converges uniformly to a function that is continuous at every point.


Let $\phi_{\mathbb{Q}\cap[0,1]}$ be the characteristic function of $\mathbb{Q}\cap[0,1]$, and $f_n=\frac{1}{n}\phi_\mathbb{Q\cap[0,1]}$. Then $f_n$ is nowhere continuous on $[0,1]$ and $f_n$ converges to $0$ uniformly.


Let $f_n(x):=1 /(1+x)^n$ for $x \in[0,1]$. Find the pointwise limit $f$ of the sequence $\left(f_n\right)$ on $[0,1]$. Does $\left(f_n\right)$ converge uniformly to $f$ on $[0,1]$ ?


$\lim f_n(0)=\lim 1=1$ and if $x>0,\lim 1 /(1+x)^n=0$. So $$ f(x):= \begin{cases}0, & 0 < x \leq 1 \\ 1,& x=0 \end{cases} $$ Since $f$ is not continuous, $\left(f_n\right)$ does not converge uniformly to $f$ on $[0,1]$


$\text { Show that } \lim \int_1^2 e^{-n x^2} d x=0 \text {. }$


$||e^{-n x^2}||_{[1,2]}=e^{-n}$. So $e^{-n x^2}$ converges to $0$ uniformly. By theorem 8.2.4 or 8.2.5, we have $ \lim \int_1^2 e^{-n x^2} d x=0$. Q.E.D.


Let $f_n(x):=x / n$ for $x \in[0, \infty), n \in \mathbb{N}$. Show that $\left(f_n\right)$ is a decreasing sequence of continuous functions that converges to a continuous limit function, but the convergence is not uniform on $[0, \infty)$.


For any fixed $x\in [0, \infty)$, $\frac{x}{n}>\frac{x}{n+1}$ and $\lim_{n\to\infty}\frac{x}{n}=0$.

Let $x_n=n$, then $f_n(x_n)=1$ is not a sequence which converges to $0$, and so the convergence is not uniform. Q.E.D.

Provided by Yunsong Wei