Proving Continuity of Functions with Contradiction

[JG89]

Homework Statement

Let f and g be continuous functions defined on all of R. Prove that if f(a) \neq g(x) for some a \epsilon R, then there is a number \delta > 0 such that f(x) \neq g(x) whenever |x-a| < \delta.

Homework Equations

I would like to please check if my proof is correct :)

The Attempt at a Solution

Suppose that f(x) = g(x) whenever |x-a| < \delta. Then by the continuity of f and g we have |f(x) - f(a)| < \epsilon whenever |x-a| < \delta and |g(x) - g(a)| < \epsilon whenever |x-a| < \delta. Since g(x) = f(x) whenever |x-a| < \delta, we can write |g(x) - f(a)| < \epsilon. Obviously the left hand side must tend to 0 as x tends to a, and so for values of g(x) > f(a), we have \lim_{x \rightarrow a} g(x) - \lim_{x \rightarrow a} f(a) = 0, implying that \lim_{x \rightarrow a} g(x) = f(a).
Now for values of g(x) < f(x) we have \lim_{x \rightarrow a} f(a) - \lim_{x \rightarrow a} g(x) = 0, implying again that \lim_{x \rightarrow a} g(x) = f(a). Since g is continuous, we must have \lim_{x \rightarrow a} g(x) = g(a), but limits are unique, so this means that f(a) = g(a), which is a contradiction. QED

 

[Unco]

Hi JG89,

I got a little lost with your arguments (does your original hypothesis hold for all delta; mixing epsilon-delta and limits can be confusing).

It is often the case in these sort of exercises that we are given a suitable "epsilon" to work with in the question. In this case, if WLOG we assume f(a)>g(a) for some a, then our epsilon might be f(a)-g(a)>0.

Then, recognising that f-g is also continuous, we may apply the epsilon-delta definition of continuity to f-g at a:

Given \varepsilon = f(a)-g(a)&gt;0 there is a \delta_\varepsilon&gt;0 such that

|x-a|&lt;\delta_\varepsilon \Rightarrow |(f(x)-g(x)) - (f(a)-g(a))| &lt; f(a)-g(a).

You continue this for yourself to obtain the desired result. These questions can often be tackled by playing around with the epsilon-delta definition in this manner.

 

[JG89]

Please see below!

 

[JG89]

Double Post

 

[JG89]

Triple Post...See below

 

[JG89]

Sorry, the latex is messing up on me!

I will reword it so it is more clear.

To prove this I will use a proof by contradiction. Assume that f(x) = g(x) whenever |x-a| &lt; \delta. Since both functions are continuous, we have |f(x) - f(a)| &lt; \epsilon whenever |x-a| &lt; \delta and |g(x) - g(a)| &lt; \epsilon whenever |x-a| &lt; \delta. Now, since we have made the assumption that f(x) = g(x) then we can rewrite |f(x) - f(a)| &lt; \epsilon as |g(x) - f(a)| &lt; \epsilon. Since this equality must be true for all epsilon greater than zero, including arbitrarily small epsilon, the left side of the inequality must tend to zero as g(x) gets closer to f(a); that is, as x approaches a. So we have \lim_{ x \rightarrow a} |g(x) - f(a)| = 0. We will separate this limit into two cases because of the absolute value bars. For the first case we have \lim_{x \rightarrow a} [g(x) - f(a)] = 0, implying that \lim_{x \rightarrow a} g(x) - \lim_{x \rightarrow a} f(a) = 0, implying that \lim_{x \rightarrow a} g(x) = \lim_{x \rightarrow a} f(a). Now, since the function is continuous we must have \lim_{x \rightarrow a} g(x) = g(a). So our limit equation turns into g(a) = f(a). Which is a contradiction because we know that f(a) \neq g(a). For the second case, where we want to evaluate the limit as x approaches a of - [g(x) - f(a)] = f(a) - g(x) it is easy to see that we will also arrive at the same contradiction. Therefore f(x) \neq g(x) whenever |x-a| &lt; \delta. QED