Show that iff is uniformly continuous on a bounded interval I, then
f is bounded on I.
let I =[a,b] a<b.
Since f is uniformly continuous on [a,b] we have :
for all ε>0 there exists a δ>0 and such that:
for all x.yε[a,b] and |x-y|<δ ,then |f(x)-f(y)|<ε........................................... ........................................1
ΝΟW subdivide [a,b] into,n subinterval of equal lengths less than or equal to δ ,as follows:
Choose .This is possible since b-a and δ are fixed positive Nos.
Choose points such that
,i = 1,2,3.............n [ the inequality follows from the choice of
Then (1) implies that:
and hence
Similarly
.................................................. ................................2
Repeating this argument for each successive subinterval up to the nth subinterval, we obtain:
.
Hence ,if we choose we obtain:
,
hence f is bounded on I
There is another proof by contradiction :
Assume that f is not bounded over [a,b],then for all ε>0 there exists an xε[a,b] and such that |f(x)|>ε .THAT implies that for all nεN ,THERE exists xε[a,b] and such that |f(x)|>n.
THAT allow us to pick ,for each n an in [a,b] and such that
THUS we have formed a sequence { } within [a,b].
BUT according to the Weistrass Bolzano theorem every bounded sequence of Nos has an accumulation point ,hence { } has an accumulation point ,vε[a,b].
That implies that there exists a subsequence of { }, { } which converges to v and such that for each n.
BUT since f is continuous in [a,b] and { } has a limit in [a,b] ,then .
THUS for every +ve No ( and thus for 1>0) there exists kεN AND such
.
Also ,since for each n , for every +ve No there exists k and .
A Contradiction and hence f is bounded in [a,b]