Consider the non linear map $U:C^0([0,1])\rightarrow C^0([0,1])$ defined by $$U(f)(x)=e^{-f(x)^2}$$ Show that $U$ is continuous.
I know various definition of continuity, like the definition in terms of limits of functions $$\lim_{x\rightarrow c}f(x)=f(c)$$ the definition in terms of limits of sequences $$\lim_{n\rightarrow\infty}x_n=c \implies \lim_{n\rightarrow\infty}f(x_n)=f(c)$$ or moreover the geometric definition $f:X\rightarrow Y$ is continuous if for every open set $V\subset Y$ the inverse image $f^{-1}(V)$ is an open subset of $X$.
Now, probably the solution is very easy but I can't understand what I have to use and where I have to start, with some formula or property.
Someone could help me?
The standard metric on $C([0,1])$ is $d(f,g) = \max_{x \in [0,1]} |f(x) - g(x)|$.
Consider the function $h(x) = e^{-x^2}$. For any two points $x,y \in \mathbb R$ you could use (for instance) the mean value theorem to find $$h(x) - h(y) = h'(z) (x-y)$$ for some point $z$ in between $x$ and $y$. Since $h'(z) = e^{-z^2}(-2z)$, it is not hard to show using the first derivative test that $h'$ is bounded above and below by $1$ and $-1$ respectively. Thus $|h'(z)| \le 1$ for all $z$ and $$|h(x) - h(y)| \le |x-y|.$$
For any point $x \in [0,1]$ you get $$|Uf(x) - Ug(x)| = |h(f(x)) - h(g(x))| \le |f(x) - g(x)| \le d(f,g).$$ This leads to $$ d(Uf,Ug) = \max_{x \in [0,1]} |Uf(x) - Ug(x)| \le d(f,g).$$ From here the continuity of $U$ is easily established using an $\epsilon$-$\delta$ argument.