Space of Lipschitz functions is complete. Show that the metric space of Lipschitz functions $f:[a, b]\rightarrow \mathbb{R}$ is complete with the following metric.
\begin{equation*}
    d(f, g)=\sup_{x\in[a, b]}|f(x)-g(x)|+\sup_{{x, y\in[a, b]},x\neq y}\frac{|(f-g)(x)-(f-g)(y)|}{|x-y|}
\end{equation*}
I have provided the proof by proving the functions have the same lipchitz constant whenever they are cauchy.
 A: The metric space of Lipschitz functions $f:[a, b]\rightarrow \mathbb{R}$ is complete with the following metric.
\begin{equation*}
    d(f, g)=\sup_{x\in[a, b]}|f(x)-g(x)|+\sup_{{x, y\in[a, b]},x\neq y}\frac{|(f-g)(x)-(f-g)(y)|}{|x-y|}
\end{equation*}
Let $f_n(x)$ be the cauchy sequence in the metric space. Since this is cauchy sequence of Lipschitz continous functions which implies they are continuous and since the space of continuous functions with the uniform metric (the first summand in our metric d) is complete we have a uniform limit $f$ which is continuous. In symbols, for given $\epsilon > 0$ there exist a $N_u$ such that for $n\geq N_u$ the following holds.
\begin{equation}
    d_u(f_n,f)=\sup_{x\in[a, b]}|f_n(x)-f(x)|<\epsilon
\end{equation}
Since, $f_n$ is cauchy we have the following, for any $\epsilon > 0$ and $n,m\geq N_0$ we have
\begin{align*}
    d(f_n, f_m)&=\sup_{x\in[a, b]}|f_n(x)-f_m(x)|+\sup_{{x, y\in[a, b]},x\neq y}\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{|x-y|}< \epsilon
\end{align*}
Let $\epsilon = 1$. Then we have $N$ such that for all $n,m\geq N$,
\begin{align*}
    \sup_{{x, y\in[a, b]},x\neq y}\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{|x-y|}&< 1 \\
    |(f_n-f_m)(x)-(f_n-f_m)(y)|&< |x-y|
\end{align*}
We claim that the the Lipchitz constants are bounded above by some real number. For any $n,m\geq N$ we have
\begin{align*}
    |f_n(x)-f_n(y)|&=|f_n(x)-f_N(x)+f_N(x)-f_N(y)+f_N(y)-f_n(y)|\\
    &= |(f_n-f_N)x-(-f_N(y)+f_n(y))+f_N(x)-f_N(y)|\\
    &\leq |(f_n-f_N)x-(f_n-f_N)(y)|+|f_N(x)-f_N(y)|\\
    &\leq |x-y|+K_N|x-y|\\
    & \leq (1 + K_N) |x-y|
\end{align*}
Let $M=\sup\{K_1, K_2, \cdots,1 + K_N\}$, then for all $f_n$ we have the same constant.
\begin{equation}
    |f_n(x)-f_n(y)|\leq M|x-y|
\end{equation}
We have a found a constant $M$ independent of $x, y$. Hence, eq.(2) holds for all $x, y$ and all of $f_n$.
Next, we show that $f$ itself is lipschitz follows from (1) and (2). Let $n\geq N_u$, and since it is convergent with respect to uniform metric.
\begin{align*}
    |f(x)-f(y)| &\leq |f(x)-f_{N_u}(x)+f_{N_u}(x)+f_{N_u}(y)-f_{N_u}(y)-f(y)| \\
    &\leq |f(x)-f_{N_u}(x)+f_{N_u}(x)+f_{N_u}(y)-f_{N_u}(y)-f(y)|\\
    &\leq |f(x)-f_{N_u}(x)|+|f_{N_u}(x)-f_{N_u}(y)|+|f_{N_u}(y)-f(y)|\\
    &\leq 2\epsilon+|f_{N_u}(x)-f_{N_u}(y)|\\
    &\leq 2\epsilon +M|x-y|
\end{align*}
As the above holds for any $\epsilon >0$ we see that $f$ is Lipschitz with constant $M$.
Now we need to prove that it is the limit of $f_n$ in our metric space too. As the first summand can be made less than any positive number we need to take care of second summand. As $m\rightarrow \infty$ we have for $\epsilon /2$ we have $n >N= \max\{N_{cauchy}, N_{uc}\}$
\begin{align*}
\lim_{m\rightarrow \infty}\frac{|(f_n-f_m)(x)-(f_n-f_m)(y)|}{|x-y|}=\frac{|(f_n-f)(x)-(f_n-f)(y)|}{|x-y|}\leq \epsilon/2
\end{align*}
Combining both the epsilons we have
$d(f_n(x), f(x))<\epsilon$ where $f$ is lipschitz function. $\blacksquare$
