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 $x$.

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    $\begingroup$ Ultimately your estimate has to reach $\varepsilon$, which is independent of $x$, so the dependence on $x$ has to be dropped sooner or later. $\endgroup$ – Michal Adamaszek Oct 21 '19 at 7:19
  • $\begingroup$ 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. $\endgroup$ – Kevin Oct 21 '19 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.

  • $\begingroup$ I think you are talking about uniformly continouous functions $\endgroup$ – David Oct 21 '19 at 7:35
  • $\begingroup$ @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. $\endgroup$ – skyking Oct 21 '19 at 8:46
  • $\begingroup$ First we pick an $x$ and $\epsilon$, then we find $\delta$ for that particular $x$ and $\epsilon$ $\endgroup$ – David Oct 21 '19 at 11:57
  • $\begingroup$ @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) $\endgroup$ – skyking Oct 21 '19 at 12:36
  • $\begingroup$ 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 $x$ $\endgroup$ – David Oct 21 '19 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 $x_0-1<x<x_0+1$ that is $|x-x_0|<1$ and in this case it is sufficient to assume


to satisfy the condition given by the definition.


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