Let $F$ be the set of all continuous functions $[0,1]\to\mathbb{R}$ and
$d_\infty(f,g):=\sup_x |f(x)-g(x)|$
$d_2(f,g):=\sqrt{\int_0^1 (f(x)-g(x))^2\, dx}$
metrics on $F$. Check which of the following functions are continuous and which are not:
a) $\operatorname{id}: (F, d_\infty)\to (F, d_2)$
b) $\operatorname{id}: (F, d_2)\to (F, d_\infty)$
c) $\operatorname{ev_0}: (F, d_\infty)\to\mathbb{R}$
d) $\operatorname{ev_0}: (F, d_2)\to\mathbb{R}$
I think, that a) and c) are continuous and b) and d) are not continuous.
A function $f: X\to Y$ between metric spaces is continuous in $x\in X$, if for every $\varepsilon >0$ it exists $\delta>0$ such that for every $x'\in X$ holds, that $d(x,x')<\delta\Rightarrow d(f(x), f(x'))<\varepsilon$.
a):
Let $f,g\in F$ and let $\varepsilon >0$ be arbitrary. Choose $\delta:=\varepsilon$.
It is $d_\infty(f,g)=\sup_{x}|f(x)-g(x)|<\delta$.
Then
$d_2(\operatorname{id}(f),\operatorname{id}(g))=d_2(f,g)=\sqrt{\int_0^1 (f(x)-g(x))^2\, dx}\leq \sqrt{\int_0^1 (\sup_x |f(x)-g(x)|)^2\, dx}<\sqrt{\int_0^1 \delta^2\, dx}=\sqrt{\delta^2}=\delta=\varepsilon$.
$\square$
c) is pretty much the same:
Let $f,g\in F$ and let $\varepsilon >0$ be arbitrary. Choose $\delta:=\varepsilon$.
It is $d_\infty(f,g)<\delta\Leftrightarrow \sup_x |f(x)-g(x)|<\delta$.
Now $d(\operatorname{ev_0}(f), \operatorname{ev_0}(g))=d(f(0), g(0))=|f(0)-g(0)|\leq \sup_x |f(x)-g(x)|<\delta=\epsilon$. $\square$
Is this correct. Could it be improved, how I write the proof down?
Next I want to show, that b) and d) are not continuous. But I am not exactly sure how to do it.
I choose some $f\in F$ and then try to show, that
$\exists\varepsilon >0\forall\delta >0\exists g\in F: d_2(f,g)<\delta\wedge d(\operatorname{ev_0}(f), \operatorname{ev_0}(g))\geq\varepsilon$
So I choose $\varepsilon$ and the function $g$ can depend on $\delta$. But I just can not find suitable choices. But I also can not proof that these functions are continuous...
For d), I thought it is best to choose $f:[0,1]\to\mathbb{R}, x\mapsto 0$ and $g: [0,1]\to\mathbb{R}, x\mapsto \delta x$
This does not work. When I choose something other than 0 for $f$ the integral
$\sqrt{\int_0^1 (f(x)-g(x))^2\, dx}$ gets out of control...
Do you have a hint?
Thanks in advance.
For both b) and d) consider the functions $f_n(x)=1-n^{2}x$ for $ 0\leq x \leq \frac 1 {n^{2}}$ and $0$ for $ x > \frac 1 {n^{2}}$. Let $f(x)=0$ for all $x$. Since $0 \leq f_n(x) \leq 1$ we get $\int f_n^{2} =\int_0^{1/(n^{2})} f_n^{2} \leq \frac 1 {n^{2}} \to 0$ so $d_2(f_n,f) \to 0$. But $f_n(0)=1$ for all $n$ so $d_{\infty} (f_n,f)$ does not tend to $0$ and $f_n(0)$ does not tend to $f(0)$.