Show that iff is uniformly continuous on a bounded interval I, then

f is bounded on I.

- Apr 15th 2009, 11:32 PMmancillaj3Show that if f is uniformly continuous on a bounded interval I, thenShow that iff is uniformly continuous on a bounded interval I, then

f is bounded on I.

- Apr 16th 2009, 05:32 AMxalk
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 - Apr 16th 2009, 09:07 AMPlato
- Apr 16th 2009, 04:09 PMxalk
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]