Hard one...

Dec 2009
1,506
434
Russia
Prove that if \(\displaystyle (f_n)^\infty_{n=1}\) is a sequence of functions that uniformly converges in [0,1] to bounded function \(\displaystyle f\), and if \(\displaystyle \varphi\) is function which defined on all \(\displaystyle \mathbb{R}\) and Lifshitz on every close interval (in other words:
for all interval \(\displaystyle [a,b]\) \(\displaystyle \exists L \geq 0\) so that \(\displaystyle y_,y_2 \in[a,b]\), \(\displaystyle |\varphi(y_1)-\varphi(y_2)|\leq L|y_1-y_2|\)) then also sequnce of functions \(\displaystyle \varphi(f_n(x))\) uniformly converges in [0,1]. What is the limit?



Thank you very much!
 
Jul 2009
555
298
Zürich
Prove that if \(\displaystyle (f_n)^\infty_{n=1}\) is a sequence of functions that uniformly converges in [0,1] to bounded function \(\displaystyle f\), and if \(\displaystyle \varphi\) is function which defined on all \(\displaystyle \mathbb{R}\) and Lifshitz on every close interval (in other words:
for all interval \(\displaystyle [a,b]\) \(\displaystyle \exists L \geq 0\) so that \(\displaystyle y_,y_2 \in[a,b]\), \(\displaystyle |\varphi(y_1)-\varphi(y_2)|\leq L|y_1-y_2|\)) then also sequnce of functions \(\displaystyle \varphi(f_n(x))\) uniformly converges in [0,1]. What is the limit?
The limit is \(\displaystyle \varphi(f(x))\).
Proof: Let \(\displaystyle \varepsilon>0\) be given. Since \(\displaystyle f_n\) converges uniformly to \(\displaystyle f\) and \(\displaystyle f\) is bounded, there exists an \(\displaystyle n_1\) such that the values of \(\displaystyle f_n(x)\) and \(\displaystyle f(x)\) for \(\displaystyle x\in[0;1]\) are contained in a compact interval \(\displaystyle [a;b]\), and hence an \(\displaystyle L\) exists, such that \(\displaystyle |\varphi(y_n)-\varphi(y)|\leq L|y_n-y|\), provided \(\displaystyle y_n,y\in[a;b]\).

Again from the uniform convergence of \(\displaystyle f_n\) to \(\displaystyle f\) it follows that there exists an \(\displaystyle n_0\), so that for all \(\displaystyle n>n_0\) and all \(\displaystyle x\in\,[0,1]\), we have that \(\displaystyle |f_n(x)-f(x)| < \frac{\varepsilon}{L}\).
Thus
\(\displaystyle |\varphi(f_n(x))-\varphi(f(x))|\leq L|f_n(x)-f(x)|<L\frac{\varepsilon}{L}=\varepsilon\), for all \(\displaystyle n>\mathrm{max}(n_0,n_1)\) and all \(\displaystyle x\in \,[0;1]\).