# Is this result correct? Limits don't depend on norm.

Let $\|\cdot\|: \mathbb R \to \mathbb R$ be some norm on $\mathbb R$ which is not the standard absolute value $|\cdot|$ we all know and love. $f: \mathbb R \to \mathbb R, L \in \mathbb R.$

Suppose that $\lim_{t \to 0} f(t) = L$ with respect to $\| \cdot \|$

This means that $\forall \epsilon > 0 \exists \delta >0: \|t\| \leq \delta \implies\|f(t)-L\| \leq \epsilon$

I would like to show that this is also true with respect to $|\cdot |$. The limit does not depend on with respect to which norm does $t$ approach zero.

What I did:

Fix $\epsilon >0$ and choose a $\delta$ such that $\|t\| \leq \delta \implies\|f(t)-f(0)\| \leq \epsilon$.

All norms are equivalent, so there is $c >0$ such that $c|t| \leq \|t\| \leq \delta$ and $c|f(t)-L| \leq \|f(t)-L\| \leq \epsilon$

So $|t| \leq \frac{\delta}{c} = \delta'$ and $|f(t)-L| \leq \frac{\epsilon}{c} = \epsilon'$

Finally, $|t| \leq \delta ' \implies\frac{\|t\|}{c} \leq\frac{\delta}{c} \implies \|t\| \leq \delta$

That implies by our initial assumption that $\|f(t)-L\| \leq \epsilon$, but then $|f(t)-L| \leq \frac{\epsilon}{c} = \epsilon'$.

Since our $\epsilon$ is as small as we would like, so is $\epsilon'$, and that concludes the proof.

Is this proof correct?

Question number 2 The differential of a function activated on unit vector $h$ is defined to be $D_f(a)h = \lim_{t \to 0} \frac{f(a+th)-f(a)}{t}$.

Suppose my proof is correct and it does not matter with respect to which norm does $t$ approach zero here. Does it matter with respect to which norm is $h$ a unit vector?

In other words- Does the differential of a function at all depend on the choice of the norms and inner products we work with? If so - why? if not - why not?

• @gimusi apologies! I just saw it now. I actually managed to solve that question on my own so I moved ahead and forgot to tend to that page. Thank you, good answer. Liked the use of the binomial theorem and little "o" notation. – Oria Gruber Feb 25 '18 at 1:25

The argument you gave contains the main idea to prove what you want to show.

But there is one point that should be revised. You say "Finally, $|t|\le\delta'\Longrightarrow \frac{\|t\|}c\le\frac\delta c$". Probably this is true in this case, but it is not obviously derived from the equations and inequalities you have. You should improve the arguments in order to prove the implication.

Spoiler: better using the existence of $c$ such that $\|t\|\le c|t|$.

For the second question: first, note that your definition does not depend on $h$ being a unit vector. Then, try to prove the following (a generalization of your first question):

Let $f:\mathbb{R}\to\mathbb{R}^n$ and let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two (equivalent) norms in $\mathbb{R}^n$. Then $\|f(t)\|_1\to0$ as $t\to0$ if and only if $\|f(t)\|_2\to0$ as $t\to0$.

Notation. $\|f(t)\|\to0$ as $t\to0$ means $$\lim_{t\to0}\|f(t)\|=0.$$

(one last commentary: $t\to0$ if and only if $|t|\to0$, so if and only if $\|t\|\to0$, where $\|\cdot\|$ is any norm in $\mathbb{R}$.)