Norm and metrics Let $E=C^1[a,b]$ and $\|\cdot\|_0$, $\|\cdot\|_1$ norms defined as $$\|f\|_0=\max_{x\in [a,b]} |f(x)|$$ and $$\|f\|_1=\max_{x\in [a,b]} |f(x)|+\max_{x\in [a,b]} |f'(x)|.$$
For $r>0$, consider the open ball with center in the origin defined by: $$B_r^0(0)=\{f \in E;\|f\|_0<r\}$$ and $$B_r^1(0)=\{f \in E;\|f\|_1<r\}.$$

*

*Prove that $B_r^1(0) \subset B_r^0(0)$.

*Prove that there is no $\epsilon>0$ such that $B_\epsilon^0 (0) \subset B_r^1(0)$.

Can I have some hints? This does not look very intuitive and I'm having some trouble to prove it.

To get 1. let $f \in B_r^1(0)$, i.e. $\lVert f \rVert_0 + \lVert f' \rVert_0 < r$. Now we conclude as norms are non-negative:
$$
\lVert f \rVert_0 \leq \lVert f \rVert_0 + \lVert f' \rVert_0 < r 
$$
So $f \in B_r^0(0)$.
To get 2. let us assume that such $\varepsilon$ exists. Then set $f_\delta: [a, b] \rightarrow \mathbb{R}$,
$$
f_\delta(x) := \frac{\varepsilon}{\sqrt{b-a +\varepsilon}} \cdot \sqrt{x-a+\delta}
$$
where $\delta \in (0, \varepsilon)$ is arbitrary. Clearly, $f_\delta \in C^1[a, b]$. Hence, for all $x \in [a, b]$:
$$
\lvert f_\delta(x) \rvert = \frac{\varepsilon}{\sqrt{b-a+\varepsilon}} \cdot \sqrt{x-a+\delta} \leq \frac{\varepsilon}{\sqrt{b-a+\varepsilon}} \cdot \sqrt{b-a+\varepsilon} = \varepsilon
$$
So $f_\delta(x) \in B_\varepsilon^0(0)$. Then, $f \in B_r^1(0)$ by assumption. So, for all $x \in [a, b]$:
$$
r > \lVert f_\delta(x) \rVert_0 + \lVert f_\delta'(x) \rVert_0 \geq \lVert f_\delta'(x) \rVert_0 = \max_{x \in [a, b]}\frac{\varepsilon}{2\sqrt{b-a +\varepsilon}} \cdot \frac{1}{\sqrt{x-a+\delta}} = \frac{\varepsilon}{2\sqrt{b-a +\varepsilon}} \cdot \frac{1}{\delta} \overset{\delta \downarrow 0}{\longrightarrow} \infty
$$
This is a contradiction.
