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Prove a function is continuous at a point (2)

  1. Mar 24, 2012 #1
    As in my previous thread we had:
    "Let f a function which satisfies $$|f(x)|\leq|x| \forall{x\in{\mathbb{R}}}$$

    Proof that is continuous at 0.
    We concluded that since f(0)=0 then we found a delta equal epsilon so $$|f(x)|≤|x|<ϵ$$.

    But now I have:

    $$\textrm{g continuous at 0 and g(0)=0}\Longleftrightarrow{\displaystyle\lim_{x \to{0}}{g(x)}=g(0)=0} $$
    $$ \left . \begin{matrix}{g(0)=0}\\{|f(x)|\leq |g(x)|}\end{matrix}\right \} \Longrightarrow{}f(0)=0$$

    And I have to prove that f is cont. at 0.

    So I have to fid a delta such that:
    $$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

    And I'm in the same problem as before, right? I just choose δ equal ε . If this is OK then I don't understand why it tells me that g is cont. at 0 because I never used that.

    Thank you!!
     
  2. jcsd
  3. Mar 24, 2012 #2

    Fredrik

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    Consider the g defined by
    $$
    g(x)=
    \begin{cases}
    0 &\text{if }x=0\\
    1 &\text{if }x\neq 0
    \end{cases}
    $$ and the f defined by
    $$
    f(x)=
    \begin{cases}
    0 &\text{if }x=0\\
    \frac 1 2 &\text{if }x\neq 0.
    \end{cases}
    $$
     
  4. Mar 24, 2012 #3
    You do use g is cont at 0 by writing
    I. if |x| < delta, then |(g(x) - g(0)| = |g(x)| < epsilon
    Because |f(x)| <= |g(x)|,
    if |x| < delta, then modify I. by substituting |f(x)| for |g(x)| to get
    if |x| < delta, then |f(x)| < epsilon

    In this case, delta = delta. In the last case, you substituted in the |x-a| < delta inequality, in this one you substituted in the epsilon's inequality.
     
  5. Mar 26, 2012 #4
    Ahmm... so what I did was right? Thanks
     
  6. Mar 26, 2012 #5

    Fredrik

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    No, because the choice δ=ε doesn't work. To "choose δ=ε" is to say that for all x such that |x|<ε, we have |f(x)|<ε. But how would you know that the latter inequality is implied by the former? You only know that |f(x)|≤|g(x)|. So for the choice δ=ε to work, you need something that proves that for all x such that |x|<ε, we have|g(x)|<ε.
     
  7. Mar 28, 2012 #6
    From ALL what I know (from the hypothesys, including g) I can say that:
    I have to a delta such that this condition is OK:

    $$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

    So if I choose epsilon equal delta:
    $$|x|< \varepsilon$$ and this implies:
    $$|f(x)|< \varepsilon$$

    EDIT: Sorry I got confused, I cannot deduce the latter inequality. But now I look at it... as Bearded Man said I have that $$|g(x)|<\varepsilon$$... so that may be useful..
     
    Last edited: Mar 28, 2012
  8. Mar 28, 2012 #7
    I think I got something, please anybody verify if this is right:

    This is what I have to proove:
    $$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

    And I know that g is cont. at 0 so:
    $$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|g(x)-0|< \varepsilon}$$
    And
    $$|f(x)|\leq |g(x)|$$

    So having that two conditions I can deduce that
    $$|f(x)|<\varepsilon$$ and therefore I can conclude that f is continuous at 0... am I missing sth?
     
  9. Mar 28, 2012 #8

    Fredrik

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    This is the right way to do it. Just make sure to mention that the delta that you use for f is the one you got from the stuff you know about g.
     
  10. Mar 28, 2012 #9
    OK, yes, I am supposing that epsilon and delta are the same numbers for g and f. Thank you very much!!!
     
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