# Show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces

Let $$\mathcal{L}$$ be the space of real-valued Lipschitz functions of order 1 defined on $$[0,1]$$. That is, the class of functions $$f$$ such that $$\sup\limits_{(x,y)\in[0,1]\times[0,1], x\neq y} \dfrac{\vert f(x)-f(y)\vert}{\vert x-y\vert} =K(f)<\infty.$$ Given $$\Vert f\Vert_1=\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f)=\Vert f\Vert + K(f)$$ is a norm on $$\mathcal{L}$$.

Show that $$\left(\mathcal{L},\Vert\cdot\Vert_1\right)$$ is a Banach spaces.

To show $$\left(\mathcal{L},\Vert\cdot\Vert_1\right)$$ is a Banach spaces, I use the definition of Banach spaces, i.e. $$\left(\mathcal{L},\Vert\cdot\Vert_1\right)$$ is complete.

To show $$\left(\mathcal{L},\Vert\cdot\Vert_1\right)$$ is complete, I will prove every Cauchy sequence on $$\mathcal{L}$$ is converge, i.e.

Let $$\{f_i\}_{i\in \mathbb{N}}$$ be the sequence in $$\mathcal{L}$$. For all $$\varepsilon>0$$ there exist $$N\in \mathbb{N}$$ such that for all natural number $$m,n>N$$ $$\Vert f_m-f_n\Vert_1<\varepsilon.$$

Proof.

Let $$\varepsilon>0$$.

\begin{align*} \Vert f_m-f_n\Vert_1 &=\sup\limits_{0\leq t\leq 1}\vert f_m(t)-f_n(t)\vert +K(f_m-f_n)\\ &\leq \sup\limits_{0\leq t\leq 1}\vert f_m(t)\vert + \sup\limits_{0\leq t\leq 1}\vert f_n(t)\vert +K(f_m-f_n)\\ \end{align*}

Now, I don't know (don't have an idea) how to conclude less than $$\varepsilon$$. Any hint to prove it?

Let $$\epsilon >0$$. There exists $$N$$ such that $$\sup _t |f_n(t)-f_m(t)| +K(f_n-f_n)<\epsilon$$ for all $$n,m >N$$. This implies that $$\sup _t |f_n(t)-f_m(t)| <\epsilon$$ for all $$n,m >N$$ and $$K(f_n-f_m)<\epsilon$$ for all $$n,m >N$$. $$(f_n(t))$$ is a Cauchy sequence so $$f(t)=\lim f_n(t)$$ exists for each $$t$$. Letting $$m \to \infty$$ in the inequality $$|f_n(t)-f_m(t)| <\epsilon$$ we get $$|f_n(t)-f(t)| \leq \epsilon$$ for $$n >N$$. Similarly, we can let $$m \to \infty$$ in $$K(f_n-f_m) <\epsilon$$ to get $$k(f_n-f) \leq \epsilon$$ for $$n>N$$. Rest of the argument should now be clear.