# Hard one...

#### Also sprach Zarathustra

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!

#### Failure

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 $$\displaystyle x\in \,[0;1]$$.

• Also sprach Zarathustra

#### Also sprach Zarathustra

Thank you so much! (Hi)