Results 1 to 4 of 4

Math Help - Show that if f is uniformly continuous on a bounded interval I, then

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    40

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by mancillaj3 View Post
    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)|<M,


    hence f is bounded on I
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,788
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by mancillaj3 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    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]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: April 3rd 2011, 05:30 AM
  2. Show that sqrt(x) is uniformly continuous on [0,infinity)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 25th 2010, 05:46 AM
  3. Replies: 3
    Last Post: March 17th 2010, 06:12 PM
  4. show this series converge uniformly to a continuous function
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 4th 2009, 12:07 AM
  5. how to show if these are uniformly continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 1st 2009, 06:43 PM

Search Tags


/mathhelpforum @mathhelpforum