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General Epsilon-Delta Proof

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Hey guys, I've been given the following [itex] \epsilon - \delta [/itex] proof question. The trouble I'm having is I'm not 100% sure what it is actually asking for, and how to go about it. The more common proofs involving two variable functions and a given limit are easy enough to do, but I'm stuck on this one.

    2. Relevant equations

    Suppose [itex] f: \left( \mathbb R ^2 , \| . \|_2 \right) \rightarrow \left( \mathbb R , | .| \right) [/itex] is continuous at [itex] \left(a,b \right) \in \mathbb R^2 [/itex]. Prove using the [itex] \epsilon - \delta [/itex] definition only, that if we define the function

    [itex] f_b : \left( \mathbb R,|.| \right) \rightarrow \left( \mathbb R,|.| \right) [/itex] s.t. [itex] f_b \left(x \right) = f \left( x,b \right), [/itex]

    then [itex] f_b [/itex] is continuous at [itex] x=a [/itex].

    3. The attempt at a solution

    Well, considering I am not actually sure what the question is acting for I can't go very far. I was assuming you would set up the usual proof as,

    Let [itex] \epsilon > 0 [/itex]
    if [itex] |(x,b) - (a,b)| < \delta [/itex]
    then [itex] |f(x,b) - (a,b)| < \epsilon [/itex]

    Possibly then, [itex] x - a < \delta [/itex]?

    Any help would be appreciated, cheers.
     
    Last edited: Aug 19, 2011
  2. jcsd
  3. Aug 19, 2011 #2

    micromass

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    Take [itex]\varepsilon >0[/itex] fixed.

    You know that f is continuous at (a,b), thus you know that there exists a [itex]\delta>0[/itex] such that for all (x,y) it holds that

    [tex]\|(x,y)-(a,b)\|_2<\delta~\Rightarrow~|f(x,y)-f(a,b)|<\varepsilon[/tex]

    Now, what you must do is to show [itex]f_b[/itex] continuous at a. Thus you must find a [itex]\delta>0[/itex] such that for all x it holds that

    [tex]|x-a|<\delta~\Rightarrow~|f(x,b)-f(a,b)|<\varepsilon[/tex]

    Can you proceed now?
     
  4. Aug 19, 2012 #3
    I am also interested in this question. I understand what is required for the proof, but I am just stuck as to how to manipulate:
    |f(x,b) - f(a,b)|
    into something that resembles δ??

    We know that:
    0 < ||(x,y) - (a,b) || < δ [itex]\Rightarrow[/itex] |f(x,y) - f(a,b)| < ε
    as the limit is equal to the value of the function at (a,b) since it is continuous there, but how can we follow on from this to show that if y is restricted to equalling b, then:
    0 < ||(x,b) - (a,b) || = ||x-a|| < δ [itex]\Rightarrow[/itex] |f(x,b) - f(a,b)| < ε
    ?? I'm really stuck and any help would be appreciated.
     
  5. Aug 19, 2012 #4

    HallsofIvy

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    How is ||(x, y)- (a, b)|| defined? (there are several equivalent definitions- which are you using?)
     
  6. Aug 20, 2012 #5
    ||(x,y) - (a,b)|| is a norm measuring the distance from (a,b) to (x,y), what do you mean how is it defined? Distance between points in R^2, euclidean norm.
     
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