Question

Please answer in the style of a formal proof and thoroughly reference any theorems, lemmas or corollaries utilized.

Let (X, d) be a metric space. Show that the set V of Lipschitz continu- ous bounded functions from X to R is a dense linear subspace of BUC(X, R). Since, in general, V #BUC(X, R), V is not a closed subset of BUC(X, R). Hint: For f EBUC(X, R) define the sequence (fr) by fn(x) = inf{f(y) + nd(x, y); Y E X}, neN. Then prove that each fn is Lipschitz continuous (with n being a Lipschitz constant) and bounded and that fn(x) + f(x) uniformly for x E X. The latter you may try by contradiction. Notice that, for each x EX, (fn(x))neN is an increasing sequence that is bounded by f(x).

BUC stands for bounded uniformly continuous

Answer 1

The solution is given below. It follows from some manipulations with limits and the definitions of uniform convergence. The explanation is given in the write up. Hope this helps.

Suppose ft BUC (X, IR) .: JMsO st. Equivalently, - Mc f(x)CM & nex. I f(x) 1cm & nEX. in let be act as defined the hinst fn We want to prove that I fn(x4) - fn (22) land(oh, X2) since we want to prove that fn is Lipschitz continuous with constant n. If x= x2, then clearly the statement holds. so, assume WLOG that fu(an) c fn (22). (For the time being also assume that in is bounded. I will prove this after proving Lipschitz continuity Let yet be site let Eso let yEX be given. best & f(") + nd(x,y) - fn(24) Le.

- fo (w) CE-Cfr9)+nd (24,9). fn (22) < fly) + nd (224) ... By definition of tu >) fa (za) - fo(su) < €+ n(d (X3,4) – d(x,y)) Let nd(x,xx) by triangle inequality. a) f(x) - false) { et nd (24,X2) & (>o. fu(x2)- fn(sc) < nd(29,22). sunce assumption was that fuld) > fn (sa). > I fon (xs) - An (34) lcndcr.X2) with constant n. =) fn is Lipschitz continuous To prove that for is bounded, observe that, inf {f(y) + nd (acy) ly ex} c f(x) and (3,0) = f(x) fu(x) < f(x) <M. Smilarly, fly) + nd(ay) > f(4) >-19 tyex. f(x) = inf {f(y) + ndCach) lyEx}, f(y) >-M. > fn (an>-M ) – Mafu(a) <ra & fn is bounded. & fn is a banded, Lipschita continuous function with constant u & HEIN.

We now need to prove that fn(a) f(x) uniformly | xe X. Obserre nat fly) + nd(x,y) = f(x) + Cuti) d(24) & YEX. = f(x) = inf {f(y) + nd(24)} e inf{f(4) +nd (x4) = fnri (?). fn (x) < from (n) & ntx = {fr} is an increasing sequence Te prere unitam convergence, Also, as seen before, fu (se) < f(x) # xxx. - Tapare lu ful x) < f(x) & nEX. hosos - To prove equality of limit under uniform convergence, it is enough to show that goren any eso, I NEIN St. VnN , and Waft, fi (2) > f(x)-& g e Let &so be given let so be sit d(x,y) < d = 1 f(x)-f(y)) se. Let NEIN sit & non, nd>2M-&. Now, let ny, N. Suppost de let xex be arbitrary. suppose yeX st dlary) Ed. a) f(y) = f(n)-& a fly) & nd(a) f(x)-&.

Suppose dan> 5. - n.d(x,y) & and > 2M-E. f(y) + n.d(x,y) fly) and of (y) + 2M-{>-M+ 2M-4 - More > f(x)-& > fly) + nd (y) > f (M) -ų. In both cases, tusN, yox, f(y) + nd (acy) > f(x)-&. - - fn(x) = inf{f(u)t na (acy) ly ex} = f(+1)-{. od sonce a was arbitrary 4 > N = ( x (7) > F(x)-8 +) (x Cao fu(x) - frol uniformly. functions are dense > Lipschitter continuous, banded in BUC CX, IR)