Show that the two norms in $E=\left\{f\in \mathcal{C}^1[0,1]:f(0)=0\right\}$ are equivalent? I can't show that the two norms defined as 
$$||f||_{\infty}=\sup_{x\in[0,1]}|f(x)+ f'(x)|$$
 and 
$$N(f)=\sup_{x\in[0,1]}|f(x)| + \sup_{x\in[0,1]}|f'(x)|$$
 are equivalent in $E=\{f\in\mathcal{C}^1([0,1]) \text{ s.t. } f(0)=0\}$.
A first inequality in one sense is trivial.
For the other inequality, I can write:
$$
f(x)=f(0)+\int_0^x f'(t) dt
$$
Which implies 
$$
 \sup_{x\in[0,1]}|f(x)| \leq  \sup_{x\in[0,1]}|f'(x)| .
$$
And then 
$$
N(f)\leq 2 \sup_{x\in[0,1]}|f'(x)| .
$$
But I can't conclude from here. 
I sincerely thank you for your help.
 A: As you have already shown, for $f \in C^{1}[0,1]$ such that $f(0) = 0$,
$$
\sup_{x \in [0,1]} |f(x)| \le \sup_{x \in [0,1]} |f'(x)|. \tag{1}
$$
We can use this to show that $\sup |f|$ is bounded by a constant times $\sup |f + f'|$, as follows:
\begin{align*}
\sup_{x \in [0,1]} |f(x)| 
&\le \sup_{x \in [0,1]} |e^x f(x)| \\
&\le \sup_{x \in [0,1]} \left| \frac{d}{dx} e^x f(x) \right| \qquad\qquad \text{(by (1), since $e^0 f(0) = 0$})\\
&= \sup_{x \in [0,1]} |e^x f(x) + e^x f'(x)| \\
&\le e \sup_{x \in [0,1]} |f(x) + f'(x)|.
\end{align*}
Now apply triangle inequality to $|f'| = |f' + f - f|$:
\begin{align*}
\sup |f'(x)| &= \sup |f'(x) + f(x) - f(x)| \\
  &\le \sup |f(x) + f'(x)| + \sup |f(x)| \\
  &\le \sup |f(x) + f'(x)| + e \cdot \sup |f(x) + f'(x)| \\
  &= (e + 1) \sup |f(x) + f'(x)|.
\end{align*}
Therefore, $\sup |f(x)|$ and $\sup |f'(x)|$ are both bounded above by a constant times $\sup |f(x) + f'(x)|$. This shows that your two norms are equivalent.
A: Two metrics on the same set are equivalent (i.e. generate the same topology )  iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.
For brevity let $\sup_{x\in [0,1]}|g(x)|=\|g\|_S$ for any continuous $g:[0,1]\to \Bbb R.$
Idea: Consider that $f(x)+f'(x)=e^{-x}(e^xf(x))'.$
(1). If $\lim_{n\to \infty}\|f_n-f\|_{\infty}=0$:
Then $\lim_{n\to \infty}\|e^{-x}(e^x (f_n(x)-f(x))'\|_S=0,$ which implies that $\lim_{n\to \infty}\|(e^x(f_n(x)-f(x))'\|_S=0,$ which, since $f_n(0)=f(0)=0,$ implies that $$\lim_{n\to \infty}\|e^x(f_n(x)-f(x))\|_S= \lim_{n\to \infty}\sup_{x\in [0,1]} |\int_0^x(e^t(f_n(t)-f(t))'dt\,|=0$$ which   implies that $$\lim_{n\to \infty}\|f_n-f\|_S=0.$$  Now $\|f'_n-f'\|_S\leq \|(f_n+f'_n)-(f+f')\|_S+\|f-f_n)\|_S=\|f_n-f\|_{\infty}+\|f_n-f\|_S, $ so we have $$\lim_{n\to \infty}\|f'_n-f'\|_S=0.$$ Since $N(f_n-f)=\|f_n-f\|_S+\|f'_n-f'\|_S,$ therefore $\lim_{n\to \infty}N(f_n-f)=0.$
(2). If $\lim_{n\to \infty}N(f_n-f)=0$:
Since $N(f_n-f)\geq \|f_n-f\|_{\infty}, $ therefore  $ \lim_{n\to \infty}\|f_n-f\|_{\infty}=0.$
