# Doubt in proof of continuity using the $\epsilon-\delta$ definition

I am learning real analysis , and started with continuity of function. I am reading the following example of proving continuity of a real function with the '$$\delta-\epsilon$$' definition

Prove $$f(x)=2x^2+1$$ is continuous using the $$\epsilon-\delta$$ definition.

Proof : Let $$x_0$$ be in $$\mathbb{R}$$ and let $$\epsilon > 0$$. We want to show that $$|f(x)-f(x_0)|<\epsilon$$ provided there exists a $$\delta>0$$, s.t $$|x-x_0|<\delta$$ .

We note $$|f(x)-f(x_0)| = |(2x^2+1)-(2x_0^2+1)|= > |2x-2x_0^2|=2|x-x_0||x+x_0|$$.We need a bound for $$|x+x_0|$$ that does not depend on x.. We notice that if $$|x-x_0|<1$$, say , then $$|x|<|x_0|+1$$ and hence $$|x+x_0|\le |x|+|x_0|<2|x_0|+1$$.

Thus we have $$|f(x)-f(x_0)| \leq 2|x-x_0|(2|x_0|+1)$$ provided $$|x-x_0|<1$$. To arrange for $$2|x-x_0|(2|x_0|+1)<\epsilon$$, it is enough to have $$|x-x_0|<\frac{\epsilon}{2(2|x_0|+1)}$$ and also $$|x-x_0|<1$$. So we put $$\delta = min(1,\frac{\epsilon}{2(2|x_0|+1)})$$. Thus we will get our desired result.

In this whole proof what I don't understand is why we need a bound for $$|x-x_0|$$ that does not depend on $$x$$. What will happen if the bound depends on $$x$$.

• Ultimately your estimate has to reach $\varepsilon$, which is independent of $x$, so the dependence on $x$ has to be dropped sooner or later. Oct 21, 2019 at 7:19
• This may be slightly against house rule, but I just wanted to say that this is an example of how to ask a good question. Up-voted.
– user284001
Oct 21, 2019 at 8:38

Since you have to given $$\epsilon$$ produce (the existence of) a $$\delta>0$$ such that $$|f(x)-f(x_0)| < \epsilon$$ whenever $$|x-x_0|<\delta$$ (that is for any such $$x$$). The $$|f(x)-f(x_0)|<\epsilon$$ inequality must hold therefore independently of $$x$$ (given that $$|x-x_0|<\delta$$).

That is eventially you have to produce a bound for $$|f(x)-f(x_0)|$$ that is independent of $$x$$ as long as $$x$$ is within some bounds.

• I think you are talking about uniformly continouous functions Oct 21, 2019 at 7:35
• @David No, it is allowed to depend on $x_0$. It mustn't depend on $x$ as soon as $|x-x_0|<\delta$, $|f(x)-f(x_0)|$ must be less than $\epsilon$ regardless of which such $x$ is choosen. Oct 21, 2019 at 8:46
• First we pick an $x$ and $\epsilon$, then we find $\delta$ for that particular $x$ and $\epsilon$ Oct 21, 2019 at 11:57
• @David No, we first pick $x_0$ and $\epsilon$ then we should be able to pick a $\delta>0$ depending on $x_0$ and $\epsilon$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ (that is for any $x$ fulfilling the last inequality) Oct 21, 2019 at 12:36
• You've just repeated exactly what I was saying. In your original answer, it could be understood that, given $\epsilon$, the same $\delta$ should work for every $x$ Oct 21, 2019 at 13:05

To comply with the definition we need that $$\delta$$ exists for any $$\epsilon$$ such that

$$\forall x,\quad|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$$

since we are interested to the continuity around $$x_0$$ we can assume wlog that $$x_0-1 that is $$|x-x_0|<1$$ and in this case it is sufficient to assume

$$\delta=\frac{\epsilon}{2(2|x_0|+1)}$$

to satisfy the condition given by the definition.