convergences of a row of functions according to a different metric What are the equivalences between the next 2 statements (with $f,(f_n)_n \in C^1([0,1])$
a) the row $(f_n)_n$ converges to f for the $d_\infty= sup\{|f(x)-g(x)||x \in [0,1]\}$
b) the row $(f_n)_n$ converges to f for the $d=\{|f(0)-g(0)|+ sup\{|f'(x)-g'(x)||x \in [0,1]\}$
I thougt that a --> b was correct and b--> a was not correct. I want to prove the first arrow and search for a counter example for the second one.
My findings:
a--> b: The $d_\infty$ gives us that the row of functions converges uniformly, this gives us that the derivatives have to be bounded. I wanted to use topological weaker so that if it converges for the $d_\infty$ it would also converge for the $d$. I wanted to do this by proving that the funtcion $g:(  C^1([0,1],d_infty)--> ( C^1([0,1]),d): x-->x is continuous. But I can't find a good estimation. Can anybody help me here?
b-->a: I think i have to prove here that the inverse of g is not continues but i can not find a counter example. Can anybody also help me with this?
Thanks in advance
 A: Consider $f_n(x)=\dfrac{x^n}{\sqrt{n}}$.
Given $\varepsilon>0$, if $n>1/\varepsilon^2$ we have that
$$\sup_{x \in [0,1]} f_n(x) = \dfrac{1}{\sqrt{n}} < \varepsilon$$
so $f_n \overset{d_\infty}{\longrightarrow} 0$.
Now, if $n>1$
$$d(f_n,0)=|f_n(0)|+\sup_{x \in [0,1]} |f_n'(x)|=\sup_{x \in [0,1]}\dfrac{n x^{n-1}}{\sqrt{n}}=\dfrac{n}{\sqrt{n}} \longrightarrow + \infty$$
so $f_n$ does not convege to $0$ on the metric $d$.
The reciprocal is indeed true by the following theorem*:

Theorem: Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for
some point $x_0 \in [a,b]$. If $\{f_n'\}$ converges uniformly on
$[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$, to a function
$f$, and for every $x \in [a,b]$ $$f'(x)=\lim\limits_{n \to \infty}f_n'(x)$$

So, if $\{f_n\}$ converges to $f \in C^1[0,1]$ on $d$, we have in particular that $\{f_n(0)\}$ converges to $f(0)$ and that $\{f_n'\}$ converges uniformly to $f'$ on $[0,1]$.
Then, by the theorem, we have that $\{f_n\}$ converges uniformly to some $g$ such that $$g'(x)=\lim\limits_{n \to \infty}f_n'(x)=f'(x)$$
So, there exists a constant $C$ such that $g(x)=f(x)+C$ for every $x \in [0,1]$, but
$$g(0)=\lim_{n \to \infty} f_n(0)=f(0)$$
and that implies that $g=f$, so $\{f_n\}$ converges uniformly to $f$ as we wanted to see.
*The theorem is Theorem 7.17 on Rudin's Principles of mathematical analysis.
