Why is a norm a continuous function? (Question about existing proof) I'm trying to follow the proof given in this answer:
https://math.stackexchange.com/a/265595/188401
I understand the proof in general, but I have a question. It's mentioned that "In this case it suffices to take δ=ε, for the following reason", but I don't see a reason provided. So why does this suffice? What happens if delta is bigger than epsilon?
Note that I don't have enough reputation to post a comment in the existing thread, hence the new question.
 A: As usual for continuity arguments, you need to be able to chose a $\delta_\epsilon >0$ for each $\epsilon>0$. In this case,  $\delta_\epsilon = \epsilon$ is a valid choice. A choice where $\delta_\epsilon$ is bigger than $\epsilon$ would not be a valid choice, but this is not a problem.
By analogy if you want to show continuity of $x \mapsto 3x$. You might say "It suffices to take $\delta = \epsilon/3$."
A: Consider the normed space $(X,||\cdot||)$
First, a norm is a function such that maps an element in $X$ to a non-negative number, i.e. $$||\cdot||:X\rightarrow\mathbb{R_{\geq0}}$$
It has the following properties:


*

*$||\alpha {\bf{x}||=|\alpha|||x||, \alpha \in \mathbb{R}, x \in} X$

*$||{\bf{x}+y||\leq ||x||+||y||\  \forall x,y\in} X$ (Triangle inequality)
Using  2, we can have the following inequality:
$$\left|||\bf{x}||-||y||\right|\leq||\bf{x-y}||---(*)$$
Therefore, ${\bf{\forall x,y}}\in X$ such that $\bf{x}$ is close to $\bf{y}$ (in the normed sense), which means right hand side of (*) is small, the left hand side will also be small. Therefore $||\cdot||$ is continuous.:)  
A: I think you're just confused about the definition of a delta-epsilon proof for continuity in general. From MathWorld in order to prove that any function is continuous:

The claim to be shown is that for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever $|x-x_0| < \delta$, then $|f(x)-f(x_0)|<\epsilon$.

Note that this definition says nothing about the relationship between $\delta$ and $\epsilon$. It's up to the person making the proof to make a relationship and thereby prove that claim above. Usually, one chooses $\delta$ to make the simplest relationship between the two variables, and $\delta$ can easily be bigger or smaller or equal to $\epsilon$.
Now in the case of the norm, when you're trying to show that $|f(x) - f(y)| < \epsilon$ when $|x - y| < \delta$ , a simple way to do so would be to keep these facts in mind:


*

*$\epsilon > 0$ (the first essential assumption in the proof)

*$|x-y| < \delta$ (the second assumption)

*$|x-y| = |y-x| = f(x-y)$ (definition of norm)

*$|x| - |y| \le |x - y|$ (reverse triangle inequality theorem)


From (3) and (4) we can see that $f(x) - f(y) \le |x -y|$, and if we work in assumption (2), we see that $f(x) - f(y) < \delta$. Using a similar proof we can show that $f(y) - f(x) < \delta$, and therefore say that $|f(x) - f(y)| < \delta$.
Keep in mind we still have no idea what $\delta$ is. As far as we know, it's just some random real number. So we have to come up with a convenient choice for $\delta$, and this is where we keep in mind what we're trying to prove ($|f(x) - f(y)| < \epsilon$) and the assumption $\epsilon > 0$.
If we simply assign $\delta := \epsilon$, then we know $\delta > 0$ and $|f(x) - f(y)| < \epsilon$. We could easily have chosen $\delta = \epsilon/3.54535$ and then we'd have $|f(x) - f(y)| < \epsilon/3.54535 < \epsilon$ and still have a valid proof, but it's unnecessary when the simpler choice ($\delta = \epsilon$) is available.
In this case, choosing some $\delta > \epsilon$ wouldn't help you prove that the norm is continuous. To see why, you basically have $|f(x) - f(y)| < \delta$, but since $\epsilon < \delta$ also, you can't really establish a relationship between $|f(x) - f(y)|$ and $\epsilon$. Not really a useful choice of $\delta$, then.
However, if you had a function $g(x) = x/7$, for example, then you could choose any value of $\delta$ where $0<\delta \le 7\epsilon$ and have a useful proof.  You can choose $\delta > \epsilon$, but whether that's a useful choice depends on the function.
