AN EXPLICIT SOLUTION OF THE LIPSCHITZ EXTENSION ...

8 ADAM M. OBERMAN
But then
D x+ 2
g(x) = g(x+ 2 ) − g(x− 1 )
x − 1 d(x + 2 , x) + d(x− 1 , x) = K C(x) +
u 2(x) − u 1 (x)
d(x + 2 , x) + d(x− 1 , x).
so we can’t have u 2 > u 1 . A similar argument establishes the reverse inequality.
So we have established (i).
2. Next we establish (ii). Let x 0 ∈ C, by (2.18), for x ∉ C, we have
|u(x) − g(x 0 )| ≤ K C (x)d(x, x 0 ) ≤ Lip(g)d(x, x 0 )
where we have used (iv). So u(x) → g(x) as C ∌ x → x 0 .
3. Next we establish (v). Without loss of generality, assume K C (x 1 ) ≤ K C (x 2 ).
Then apply (2.19) to get
and apply (2.18) with x = x 1 and y = x + 2
g(x + 2 ) = u C(x 2 ) + K C (x 2 )d(x 2 , x + 2 )
to get
g(x + 2 ) − u C(x 1 ) ≤ K C (x 1 )d(x + 2 , x 1)
combine the two previous equations to get
u C (x 2 ) − u C (x 1 ) ≤ K C (x 1 )d(x + 2 , x 1) − K C (x 2 )d(x + 2 , x 1)
≤ K C (x 1 )(d(x + 2 , x 1) − d(x + 2 , x 1)) ≤ K C (x 1 )d(x 1 , x 2 )
which gives (2.15) as desired. Finally, applying (iv) gives (v).
4. The main assertion of item (vi) follows from (2.15). The equality conditions
follow from (2.19),(2.20).
u
K 1
K 2
x 1
x 2
C
x
Figure 3. Diagram to illustrate Lipschitz continuity of K C (x)
away from D.
5. Next we establish (2.16), follow Figure 3. First assume u C (x 1 ) ≤ u C (x 2 ).
Then using (2.19) and the assumption, we have
next, use (2.18) to get
K C (x 2 ) = g(x+ 2 ) − u C(x 2 )
d(x + 2 , x 2)
≤ g(x+ 2 ) − u C(x 1 )
d(x + 2 , x 2)
g(x + 2 ) − u C(x 1 ) ≤ K C (x 1 )d(x + 2 , x 1)