Let a sequcence of continuous function $f_{n}(x)$ be defined by $[0,1]$ is uniformly convergence constant function (ex.$f_{n}(0)=f_{n}(1)=0$)
If $f:[0,\infty ]\rightarrow \mathbb{R}$ is a function defined by
$$f(x)=f_{n}(x-n),\text{ for } x\in [n,n+1]$$
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Show that $f$ is uniformly continuous.
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If $f_{n}$ is pointwise convergent, is $f$ uniformly continuous?
I think that
first, Since $f_{n}(x)$ is uniformly convergent, constant function f is continuous
And $\exists \varepsilon >0, x\in [0,1], n,m\geq N \Rightarrow
\left \| f_{n}(x)- f_{m}(x)\right \|<\frac{\varepsilon}{3}$.
So we have $\exists k \in \mathbb{N}$ such that $\frac{1}{k}<\delta$
$$\left \langle f_{1}(\frac {0}{k}-n) \right \rangle, \left \langle f_{2}(\frac {1}{k}-n) \right \rangle, \left \langle f_{3}(\frac {2}{k}-n) \right \rangle,,,,,,\left \langle f_{n}(\frac {k}{k}-n) \right \rangle$$
then, there is satisfying $\left | f(\frac {i}{k}-n) \right |< \varepsilon$
such that i={0,1,2,3…..k},
also $\exists i'\in i$ $$n\leq x\leq n+1 \Rightarrow (x+\frac {i'}{k})-n\leq (1-\frac {i'}{k})+\frac {i'}{k}< \delta$$
So, $\left | f(x)-f(\frac {i}{k}-n) \right |\leq \left | f(x)-f_{n}(x) \right |+ \left | f_{n}(x)-f_{n}(\frac {i}{k}-n) \right ||+ \left | f_{n}(\frac {i}{k}-n)-f(\frac {i}{k}-n) \right |< \frac {\varepsilon}{3}+\frac {\varepsilon}{3}+\frac {\varepsilon}{3}=\varepsilon$
hence, f is uniformly continuous.
I need for your help. Thank you for reading for my problem
Best Answer
$f$ is continuous by construction. Since $f$ is "piecewise" defined by continuous functions, and taking the value $0$ at the boundaries of these intervals.
For uniform continuity, let $\epsilon>0$ be given, the idea is break up the domain into $[0, N]$ and $(N ,\infty)$ for some $N\in\mathbb{N}$ such that for every $n \geq N$, we have $|f_n(x) - f_N(x)|\leq \frac{\epsilon}{3}$ for each $x\in [0,1]$.
Now $f$ is uniformly continuous on $[0,N]$, there exists a $\delta_1$ such that for each $x\in [0,N]$, whenever $|x-y|\leq \delta_1$, we have $|f(x)-f(y)|\leq \epsilon$.
On $(N,\infty)$, we first look at $x\in (N, N+1]$, there exists a $\delta_2$ such that whenever $|x-y|\leq \delta_2$, we have $|f(x) - f(y)| = |f_N(x-N)-f_N(y-N)| \leq \frac{\epsilon}{3}$. Now obverse that for every $x\in (N,\infty)$, whenever $|x-y|\leq\delta_2$ we have
$$|f(x) - f(y)| = |f_n (x-n) - f_n(y-n)| \\= |f_n (x-n) - f_N(x-n) + f_N(x-n) - f_N(y-n) + f_N(y-n) - f_n(y-n)| \\\leq |f_n (x-n) - f_N(x-n)| + |f_N(x-n) - f_N(y-n)| + |f_N(y-n) - f_n(y-n)| \leq \epsilon.$$
Define $\delta = \min\{\delta_1, \delta_2\}$, we have shown that for each $x\in \mathbb{R}^+$, whenever $|x-y|\leq\delta$ we have $|f(x)-f(y)|\leq \epsilon$.