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

  1. #1
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    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.
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  2. #2
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    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 $\displaystyle n\geq\frac{b-a}{\delta}$.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
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  3. #3
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    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 $\displaystyle 1 > 0 \Rightarrow \quad \left( {\exists \delta > 0} \right)\left[ {\left\{ {x,y} \right\} \subset [a,b] \wedge \left| {x - y} \right| < \delta \Rightarrow \left| {f(x) - f(y)} \right| < 1} \right]$.
    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 $\displaystyle \left| {f(y) - f(x_j )} \right| < 1 \Rightarrow \quad \left| {f(y)} \right| \leqslant 1 + \left| {f(x_j )} \right| \leqslant B$.
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    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]
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