A norm for Lipschitz-continuous functions I am a first year undergrad math student, and I am struggling with a proof. Let the set $C_{\text{Lip}}:= \left \{ f:\mathbb{R}\rightarrow \mathbb{R}: f \text{ is Lipschitz continuous} \right \}$ be a Vectorspace over $\mathbb{R}$. For $f \in C_{\text{Lip}}$ we define 
$$
c_f:=\sup\left \{ \frac{\left |f(x)-f(y)  \right |}{\left |x-y  \right |} : x,y \in \mathbb{R}, x\neq y \right \}.
$$
Show that:
$$
\left \| f \right \|:=\left | f(0) \right |+c_f
$$
is a norm on $C_{\text{Lip}}$.
What is the best way to go about this? Any help is appreciated.
Cheers
@TBrendle:
Tanzania, in my home country brother!
 A: When one has to show that something is a certain mathematical object, there is only one way to go: to verify the definition. So, if you want to show the assigned function is a norm, you have to verify the axioms for a norm.


*

*$\lVert f \rVert \geq 0$. Obvious, since $\lVert f \rVert$ is sum of two nonnegative contributions.

*$\lVert f \rVert = 0 \iff f = 0$. Suppose $f = 0$. Then
\begin{equation}
\lVert f \rVert = \lvert f(0) \rvert + c_f = 0,
\end{equation}
since $f(x) = 0$ for all $x$ in the domain. Note that $c_f$ is well defined. Conversely, suppose $\lVert f \rVert = 0$. Since it is sum of two nonnegative contributions, this is possible if and only if $f(x) = 0$ for all $x$ in $\mathbb R$.

*$\lVert a f \rVert = \lvert a \rvert \lVert f \rVert$ ($a \in \mathbb R$). Obvious.

*Triangle inequality. Let $f,g \in C_{Lip}$. Then 
\begin{equation}
\lVert f + g \rVert = \lvert (f + g)(0) \rvert + c_{f+g} \leq \lvert f(0) \rvert + \lvert g(0) \rvert + c_f + c_g = \lVert f \rVert + \lVert g \rVert.
\end{equation}


The only non completely trivial passage is $c_{f+g} \leq c_f + c_g$. We have
\begin{equation}
\begin{split}
c_{f+g} &= \sup\left\{ \frac{\lvert (f + g)(x)-(f + g)(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& =\sup \left\{ \frac{\lvert ( f(x) - f(y) ) + (g(x)-g(y)) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& \leq \sup\left\{ \frac{\lvert (f(x)- f(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} + \sup\left\{ \frac{\lvert g(x)- g(y) \rvert}{\lvert x - y \rvert} : x,y \in \mathbb R , x \neq y \right\} \\
& = c_f + c_g.
\end{split}
\end{equation}
Hence the result.
A: Here's how it starts, pretty much. It's actually pretty straightforward.
$\lVert f\rVert$ is obviously non-negative, $\lVert \mathbf 0 \rVert=0$, and if $f\ne\mathbf0$ then either … or … must be non-zero.
\begin{align*}
\lVert f+g\rVert &= |f(0)+g(0)|+\sup\left\{\frac{|f(x)-f(y)+g(x)-g(y)|}{|x-y|}:x\ne y\right\}\\
&\le \cdots
\end{align*}
Edit: I actually forgot, momentarily, about the multiplicative rule, but that's trivial here.
