# Show that if f is uniformly continuous on a bounded interval I, then

• Apr 15th 2009, 11:32 PM
mancillaj3
Show that if f is uniformly continuous on a bounded interval I, then
Show that if
f is uniformly continuous on a bounded interval I, then

f is bounded on I.
• Apr 16th 2009, 05:32 AM
xalk
Quote:

Originally Posted by mancillaj3
Show that if

f 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 $n\geq\frac{b-a}{\delta}$.This is possible since b-a and δ are fixed positive Nos.

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

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

Then (1) implies that:

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

Similarly

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

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

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

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

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

hence f is bounded on I
• Apr 16th 2009, 09:07 AM
Plato
Quote:

Originally Posted by mancillaj3
Show that if
f is uniformly continuous on a bounded interval I, then
f is bounded on I.

Here is a different proof.
From uniformly continuity $1 > 0 \Rightarrow \quad \left( {\exists \delta > 0} \right)\left[ {\left\{ {x,y} \right\} \subset [a,b] \wedge \left| {x - y} \right| < \delta \Rightarrow \left| {f(x) - f(y)} \right| < 1} \right]$.
There is a finite collection such $\left[ {a,b} \right] \subseteq \bigcup\limits_{k = 1}^n {\left( {x_k - \delta ,x_k + \delta } \right)} ,~x_k \in \left[ {a,b} \right]$.
Define $B = \sum\limits_{k = 1}^n {\left| {f(x_k )} \right|} + 1$.

If $y \in \left[ {a,b} \right] \Rightarrow \quad \left( {\exists j} \right)\left[ {y \in \left( {x_j - \delta ,x_j + \delta } \right)} \right]$.
This means $\left| {f(y) - f(x_j )} \right| < 1 \Rightarrow \quad \left| {f(y)} \right| \leqslant 1 + \left| {f(x_j )} \right| \leqslant B$.
• Apr 16th 2009, 04:09 PM
xalk
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 $x_{n}$ in [a,b] and such that $|f(x_{n})|>n$

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

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

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

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

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

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

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

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