Prove that $g\circ f: E \to \mathbb{R^p}$ is a Lipschitz continuous function and that $Lip(g\circ f) \leq Lip(g)Lip(f)$. 
Let $E \subseteq \mathbb{R^n}$ and $f: E \to \mathbb{R^m}$ be a Lipschitz continuous function. We define $Lip(f)$ as the infimum of the Lipschitz's constants of $f$. Now, let $F \subseteq \mathbb{R^m}$ and $g: F \to \mathbb{R^p}$ such that $f(E) \subseteq F$ and that g is a Lipschitz continuous function. Prove that $g\circ f: E \to \mathbb{R^p}$ is a Lipschitz continuous function and that $Lip(g\circ f) \leq Lip(g)Lip(f)$. 

My attempt: We have that the composition of Lipschitz continuous functions is a Lipschitz continuous function, since $\|g(f(x)) - g(f(y)) \| \leq K_2  ||f(x) - f(y)\| \leq K_1K_2 \|x-y\|$
But how do I prove the inequality? Any help to finish this proof would be appreciated!
 A: First notice that for $f:E \to F$ we always have $|f(x)-f(y)|\leq \mathrm{Lip}(f)|x-y|$. Indeed, for $x\neq y$ we have 
$$\frac{|f(x)-f(y)|}{|x-y|} \leq k$$
for all Lipschitz constants $k$. Since $\mathrm{Lip}(f)$ is the infimum of all such constants $k$ we have:
$$\frac{|f(x)-f(y)|}{|x-y|} \leq \mathrm{Lip}(f).$$
Therefore, $|f(x)-f(y)| \leq \mathrm{Lip}(f)|x-y|$ for $x\neq y$. But, this formula is also true for $x=y$, because both sides of inequality became $0$ in  this case. So, $\mathrm{Lip}(f)$ is the smallest Lipschitz constant, it is the minimum of the set of all Lipschitz constants.
Now, by your argument taking $k_1 = \mathrm{Lip}(f)$ and $k_2 = \mathrm{Lip}(g)$ we have:
$$|g\circ f(x)-g\circ f(y)| \leq \mathrm{Lip}(f)\mathrm{Lip}(g)|x-y|.$$
Since, $\mathrm{Lip}(g \circ f)$ is the smallest Lipschitz constant for $g \circ f$ we obtain
$$\mathrm{Lip}(g \circ f) \leq \mathrm{Lip}(f)\mathrm{Lip}(g).$$
A: Since $K := \mathrm{Lip}(g \circ f)$ is the infimum of all Lipschitz constants, if $\lVert g \circ f(x) - g \circ f(y)\rVert \leq L\lVert x - y\rVert$, then $K \leq L$. Since $K_1K_2$ is a possible candidate for $L$, $K \leq K_1K_2$.
