Question: Let $\mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm $$\|f\| = \sup\left\{ \frac{|f(x)-f(y)|}{\|x-y\|}:x,y\in X \right\}.$$ Show that $\mathcal{L}$ is a Banach space.
My attempt:
Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $(\mathcal{L},\|\cdot\|).$ Define $f:X\to \mathbb{R}$ by $$ f(x) = \lim_{n\to\infty} f_n(x). $$ We claim that $f$ is well-defined, that is, the limit exists. Fix $x\in X\setminus \{0\}$ and $\epsilon>0.$ Since $(f_n)_{n=1}^\infty$ is Cauchy, there exists $N\in \mathbb{N}$ such that for any $m,n\geq N,$ $$ \|f_m-f_n\| <\frac{\epsilon}{\|x\|}. $$ Note that $$ \|f_m-f_n\| = \sup_{x,y\in X} \frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{\|x-y\|} \geq \frac{|f_m(x) - f_n(x)|}{\|x\|}. $$ So we have $|f_m(x) - f_n(x)| < \epsilon.$ Therefore, $(f_n(x))_{n=1}^\infty$ is Cauchy in $\mathbb{R}.$ Since $\mathbb{R}$ is complete, so $\lim_{n\to\infty} f_n(x)$ exists, that is, $f$ is well-defined.
We claim that $f\in \mathcal{L}.$ By construction, $f(0) = 0.$ Since $(f_n)_{n=1}^\infty$ is Cauchy in $(\mathcal{L},\|\cdot\|),$ therefore it is bounded, that is, there exists $B>0$ such that $\|f_n\|\leq B$ for all $n\geq 1.$ Since $$f = f_N + (f-f_N),$$ so for any $x,y\in X,$ \begin{align*} |f(x)-f(y)| & \leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \\ & \leq \|f_N\|\| x-y\| + \|f-f_N\| \|x-y\| \\ & \leq \|f_N\|\| x-y\| + \|x-y\| \\ & = (\|f_N\|+1)\|x-y\|. \end{align*} Therefore, $\|f\|$ is finite and hence $f\in \mathcal{L}.$
Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(\mathcal{L},\|\cdot\|).$ Indeed, for every $\epsilon>0,$ by Cauchyness of $(f_n),$ there exists $N\in\mathbb{N}$ such that for any $m,n\geq N,$ $$ \|f_n-f_m\|<\frac{\epsilon}{2}. $$ Then $$ \|f-f_n\| = \|\lim_{m\to\infty} f_m - f_n\| = \lim_{m\to\infty} \|f_m-f_n\| < \epsilon. $$ We conclude that $\mathcal{L}$ is complete.
Is my proof above correct?