Consider $E$ the space of the functions $f: [0,1] \to \mathbb{R}$ such that $f(0) = 0$ and $f$ satisfies a Lipschitz condition. We define two norms: $$\|f\| = \sup_{x \in [0,1]} |f(x)|$$
and $$\langle \langle f \rangle \rangle = \sup_{x \in ]0,1]} \frac{|f(x)|}{x}$$
The first part asks to prove that $\langle \langle \cdot \rangle \rangle $ is strictly finer than $\| \cdot \|$. This I managed to do. But I found the second part difficult. Let $\varphi \in E$, of class $C^1$, such that $\varphi '(0) = 0$. Let $F \subset E$ the set of functions $f$ such that $|f(x)| \leq |\varphi (x)| \quad \forall x \in [0, 1]$ We have to prove that the induced metrics on $F$ are equivalent.
My attempt : It should suffice to prove that $\langle \langle \cdot \rangle \rangle$ and $\| \cdot \|$ are equivalent. Because of the first part, we only have to find a constant $k$ such that $\langle \langle f \rangle \rangle \leq k \|f\| \quad \forall f \in F$. By the definition of $\varphi '(0)$, I get $\lim_{x \to 0} \frac{\varphi (x)}{x} = 0$ Joining this with the condition $|f(x)| \leq |\varphi (x)|$, I got $\lim_{x \to 0} \left|\frac{f(x)}{x}\right| = 0 $
Since $f$ and $\varphi$ satisfy Lipschitz conditions, exists $k_1, k_2 \in \mathbb{R}$ such that $|f(x)| \leq k_1 |x|$ and $|\varphi (x)| \leq k_2 |x|$, for every $x \in [0,1]$ But I can't join all this information. My intuition says that the constant we're looking for is related to the Lipschitz constant for $\varphi$, since it is fixed. Thanks in advance!
In general there is no constant $k$ such that $\langle \langle f \rangle \rangle \leq k \|f\| $ for all $f \in F $. For example, take $\varphi(x)=x^2$ and consider the functions $f_n$ defined by $$f_n(x) = \max(x^2, 1/n)$$ Clearly, $f_n \in F$ and $\|f_n\| = 1/n$. On the other hand, $$\langle \langle f_n \rangle \rangle \ge \frac{f_n(1/\sqrt{n})}{1/\sqrt{n}} = \frac{1}{\sqrt{n}}$$ As $n\to\infty$, the ratio $\langle \langle f_n \rangle \rangle / \|f\|$ grows to infinity.
However, the metrics are equivalent in a weaker sense: for every $\epsilon>0$ there is $\delta>0$ such that if $\|f\|<\delta$, then $\langle \langle f \rangle \rangle\le \epsilon $. To show this, introduce $\psi(x) = |\varphi(x)|/x$ and observe that $\psi(0+)=0$. Therefore, given $\epsilon$, we can choose $\tau\in (0,1)$ such that $\psi<\epsilon$ on $(0,\tau)$. Let $\delta = \tau \epsilon$.
If $\|f\|<\delta$, then $|f(x)|/x \le \psi(x) < \epsilon$ on $(0,\tau)$, and also $|f(x)|/x < \delta/\tau = \epsilon$ on $(\tau,1]$.