Suppose you had some arbitrary function [itex]f : R^n \to R^p[/itex] and [itex]x \in R^n[/itex]. You want to know if it's continuous, so you do some epsilon-delta to find out for sure. However, only the most simple functions permit this without some extra restrictions.(adsbygoogle = window.adsbygoogle || []).push({});

Consider [itex]f(x) = x^2[/itex]. To show that [itex]|x - a| < \delta \rightarrow |f(x) - f(a)| < \epsilon[/itex], you'd have to break up [itex]|x^2 - a^2|[/itex] into [itex]|x + a||x - a|[/itex]. The next step is something I'm somewhat concerned about, but not even for this particular function. The technique I want to illustrate ought to work for higher powers. If it's a correct technique, it ought to work for the general case.

In general, proofs of continuity for anything but linear combinations would require that you make an assumption [itex]|x - a| < 1[/itex] and then [itex]|x| < 1 + |a|[/itex], by the triangle inequality. In this case, we took the witnessing constant 1 to create the coefficient [itex](|1 + |a| + a|)[/itex] to be sure that this neighborhood [itex](|1 + |a| + a|)|x - a| < \epsilon[/itex] holds for the delta we wish to find.

Then when you get around to finding epsilon, you take delta as the min of whatever epsilon and 1, or [itex] \delta = \min( 1, F(\epsilon) ) [/itex] where [itex]F(\epsilon)[/itex] is a function of your constant a and any witnessing constants.

My question is: Will this always work? Can I just introduce witnessing constants to replace x whenever I don't want it? (Provided I maintain the inequality.) Will you ever run into trouble with this as a general strategy? I've seen this in all the books I've come across, so I feel the answer is yes, but having someone more experienced will help solidify that.

Surely I will need more advanced techniques for higher level courses, but is there anything I should watch for at my current level (intro analysis course)?

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# Proper handling of witnessing constants in epsilon-delta proofs

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