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

f is bounded on I.

- Apr 15th 2009, 10:32 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, 04:32 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 $\displaystyle n\geq\frac{b-a}{\delta}$.This is possible since b-a and δ are fixed positive Nos.

Choose points $\displaystyle x_{o}=a,x_{1},x_{2}.............x_{n}=b$ such that

$\displaystyle x_{i}-x_{i-1} =\frac{b-a}{n}\leq\delta$ ,i = 1,2,3.............n [ the inequality $\displaystyle x_{i}-x_{i-1}\leq\delta$ follows from the choice of $\displaystyle =n\geq\frac{b-a}{\delta}$

Then (1) implies that:

$\displaystyle x\in[x_{o},x_{1}]\Longrightarrow|f(x)|< |f(x_{o}| +\epsilon$

and hence $\displaystyle |f(x_{1})|<|f(x_{o})|+\epsilon$

Similarly

$\displaystyle x\in[x_{1},x_{2}]\Longrightarrow |f(x)|< |f(x_{1})| +\epsilon<|f(x_{o})| + 2\epsilon$.................................................. ................................2

Repeating this argument for each successive subinterval up to the nth subinterval, we obtain:

$\displaystyle x\in[x_{n-1},x_{n}]\Longrightarrow |f(x)|< |f(x_{o})| + n\epsilon$.

Hence ,if we choose $\displaystyle M\geq |f(x_{o})| + n\epsilon$ we obtain:

$\displaystyle x\in[a,b]\Longrightarrow |f(x)|<M$,

hence f is bounded on I - Apr 16th 2009, 08:07 AMPlato
Here is a different proof.

From uniformly continuity $\displaystyle 1 > 0 \Rightarrow \quad \left( {\exists \delta > 0} \right)\left[ {\left\{ {x,y} \right\} \subset [a,b] \wedge \left| {x - y} \right| < \delta \Rightarrow \left| {f(x) - f(y)} \right| < 1} \right]$.

There is a finite collection such $\displaystyle \left[ {a,b} \right] \subseteq \bigcup\limits_{k = 1}^n {\left( {x_k - \delta ,x_k + \delta } \right)} ,~x_k \in \left[ {a,b} \right]$.

Define $\displaystyle B = \sum\limits_{k = 1}^n {\left| {f(x_k )} \right|} + 1$.

If $\displaystyle y \in \left[ {a,b} \right] \Rightarrow \quad \left( {\exists j} \right)\left[ {y \in \left( {x_j - \delta ,x_j + \delta } \right)} \right]$.

This means $\displaystyle \left| {f(y) - f(x_j )} \right| < 1 \Rightarrow \quad \left| {f(y)} \right| \leqslant 1 + \left| {f(x_j )} \right| \leqslant B$. - Apr 16th 2009, 03:09 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 $\displaystyle x_{n}$ in [a,b] and such that $\displaystyle |f(x_{n})|>n$

THUS we have formed a sequence {$\displaystyle x_{n}$} within [a,b].

BUT according to the Weistrass Bolzano theorem every bounded sequence of Nos has an accumulation point ,hence {$\displaystyle x_{n}$} has an accumulation point ,vε[a,b].

That implies that there exists a subsequence of {$\displaystyle x_{n}$}, {$\displaystyle y_{n}$} which converges to v and such that $\displaystyle |f(y_{n})|>n$ for each n.

BUT since f is continuous in [a,b] and {$\displaystyle y_{n}$} has a limit in [a,b] ,then $\displaystyle \lim_{n\rightarrow\infty}{f(y_{n})} = f(v)$.

THUS for every +ve No ( and thus for 1>0) there exists kεN AND such

$\displaystyle |f(y_{k})-f(v)|< 1\Longrightarrow |f(y_{k})| <|f(v)| + 1$.

Also ,since for each n $\displaystyle |f(y_{n})|>n$, for every +ve No there exists k and $\displaystyle |f(y_{k})|> |f(v) + 1$.

A Contradiction and hence f is bounded in [a,b]