# the space of lipschitz function is complete with respect to some norm

Let $V$ be the space of real valued lipschitz functions over $[a,b]$,we define:

$M_f=sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|}$

and lipschitz norm:

$||f||_{Lip}=|f(a)|+M_f$

prove that $V$ with lipschitz norm is a complete normed vector space.

it is easy to see $V$ is a vector space,but what about the completeness?

is there any hint?

thank you very much.

You should start from a hypothetical Cauchy sequence of functions (Cauchy in the sense of the newly defined norm). Prove that the values of that function at the left end $a$ converge as real numbers, and then prove that the functions converge pointwise at any $x\in[a,b].$ This allows you to write $f(x)$ for the pointwise limit. Now establish that the sequence actually converges to $f$ in the sense of the new norm.
Explicitly, let $\epsilon>0$ be given. Pick $N$ such that for $m,n\geq N,$ $M_{f_m-f_n}<\epsilon/4.$
For every $x,y\in[a,b]$ choose $m_{x,y}\geq N$ such that $|f_{m_{x,y}}(x)-f(x)|<\epsilon|x-y|/4$ and $|f_{m_{x,y}}(y)-f(y)|<\epsilon|x-y|/4.$ Then
\eqalign{ \forall x,y\in[a,b]:&|(f(x)-f_n(x))-(f(y)-f_n(y))|\\ &\leq |f(x)-f_{m_{x,y}}(x)|+|f_{m_{x,y}}(x)-f_n(x)+f_n(y)-f_{m_{x,y}}(y)| +|f_{m_{x,y}}(y)-f(y)| \\ &\leq \epsilon|x-y|. }
• I showed for every $x$ $f_n (x)$ is cauchy, so the limit is function $g$,now I must say g is the limit of $f_n$'s with lip norm.isn't it true? – user115608 Dec 13 '15 at 15:44
• It doesn't! I only needed pointwise convergence to be able to legitimately write the expression $f(x).$ Convergence in the Lipschitz norm follows from the Cauchy nature of the sequence. – Justpassingby Dec 13 '15 at 15:44