Dgdrxrt

In my analysis textbook, we have a proof for the following theorem:

Taylor's limit-formula in k dimensions

Let $f:Omegatomathbb R$ be a $C^2$ function in the open set $Omega subseteq mathbb R^k$, and let $a in Omega$. Then we have
$$f(a + Delta x) = f(a) + nabla f(a) cdot Delta x + frac12Delta x^Tcdot D^2f(a) cdotDelta x + o(||Delta x||^2)$$
in the limit as $Delta x to 0$.

I'll reproduce our definition for small-o and part the relevant part of the proof of this theorem.

Definition: Small-o

Let $A subseteq mathbb R^k$ be a set, and let $f : A to mathbb R^m$ and $g : A to mathbb R^n$ be two functions. We say that $f = o(g)$ in the limit $x to a$ if there exists an epsilon-function $varepsilon : A to mathbb R^m$ such that
$$f(x) = varepsilon(x) cdot ||g(x)|| quad textfor all $x in A$.$$

An epsilon-function is just a function that goes to zero as $x to a$. Now, our proof for the Taylor theorem goes something like this:

Proof

Choose $varepsilon > 0$ arbitrarily. Now choose $delta > 0$ such that bunch of things are satisfied.

After a bunch of manipulations, we have shown that:
$$|text| leq kvarepsilon ||Delta x||^2$$
when $||Delta x|| < delta$, thus the expression is clearly $o(||Delta x||^2)$.

Now, of course $kvarepsilon$ is not an epsilon-function, as it is a positive real number, and thus does not have a limit of zero as $Delta x to 0$, so the proof does not actually directly use the definition of small-o.

While I can see how the fact that we have chosen $varepsilon$ arbitrarily should allow us to construct some sort of epsilon-function here, they haven't done so, and constructing this epsilon-function is, in my opinion, rather complicated. The construction I came up with is:

My construction of the epsilon-function
In the proof we are able to choose a $delta$ for every $varepsilon$. This means we could create a function $g : mathbb R_+ to mathbb R_+$ defined by $g(varepsilon) = textthe $delta$ for that $varepsilon$$, and since decreasing a $delta$ still yields a valid one, we can choose the $delta$s such that $g$ is increasing and has a limit of zero as $varepsilon to 0_+$.

Now we can create a function $h : mathbb R_+ to mathbb R_0,+$ defined by $h(delta) = inf_varepsilon in mathbb R_+delta leq g(varepsilon)$. (This is just the inverse of $g$ except it's also defined if $g$ is not surjective.) Notice that $h$ is also increasing and has a limit of zero as $delta to 0_+$.

Let's look at $xi(Delta x) = kh(2||Delta x||)$. Notice that $xi$ is an epsilon-function. By construction of $h$, if we choose $varepsilon = h(2||Delta x||)$ we get a $delta$ such that $||Delta x|| < 2||Delta x|| leq delta$. This means that since $$kvarepsilon ||Delta x||^2 = kh(2||Delta x||)||Delta x||^2 = xi(Delta x)||Delta x||^2,$$ and since $||Delta x|| < delta$, we have found an epsilon-function $xi$ such that $$|text| leq xi(Delta x)||Delta x||^2$$ for small enough $Delta x$.

Thus we have showed the thing is less than a $o(||Delta x||^2)$ function, and thus is also $o(||Delta x||^2)$.

The question is: Is there a simpler way to change the proof such that it doesn't handwave the fact that it is $o(||Delta x||^2)$.

One thing I tried is to add another condition to the selection of $delta$, namely $delta < varepsilon$, hoping I could use $||Delta x|| < delta$ to replace the $varepsilon$ with a $||Delta x||$, which is an epsilon-function, but the inequality is then $||Delta x|| < varepsilon$, while the thing we would need is $||Delta x|| geq varepsilon$.