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About LinkBacksShow that iff is uniformly continuous on a bounded interval I, then
f is bounded on I.
let I =[a,b] a<b.Originally Posted by mancillaj3
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 pointssuch that
,i = 1,2,3.............n [ the inequality
follows from the choice of
Then (1) implies that:
![x\in[x_{o},x_{1}]\Longrightarrow|f(x)|< |f(x_{o}| +\epsilon](http://latex.codecogs.com/png.latex?x\in[x_{o},x_{1}]\Longrightarrow|f(x)|< |f(x_{o}| +\epsilon)
and hence
Similarly
.................................................. ................................2
Repeating this argument for each successive subinterval up to the nth subinterval, we obtain:
.
Hence ,if we choosewe 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 anin [a,b] and such that
THUS we have formed a sequence {
BUT according to the Weistrass Bolzano theorem every bounded sequence of Nos has an accumulation point ,hence {
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 {.
THUS for every +ve No ( and thus for 1>0) there exists kεN AND such
.
Also ,since for each n.
A Contradiction and hence f is bounded in [a,b]
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