# Taylor theorem & differentiability

Problem: $$U\text{ (open) } \in R^n$$. $$F: U \rightarrow R$$ is $$C^2.$$ $$A \text{ (closed rectangle) } \in U$$. Show that there exists $$\varepsilon$$ in $$\mathbb R^n$$ so that $$\left|\frac{f(x+h) - f(x) - Df(x)(h)}{\|h\|^2}\right| \leq \varepsilon \quad \forall x \in A.$$

Proof:

By Taylor's theorem we can simplify the second degree approximation as: $$p_{f,x}^2 = f(x) + Df(x)(h) + \sum_{I \in \mathcal{I}_n^2} D_I f(x) h^I$$

So we have: $$\lim_{h \rightarrow 0}\frac{f(x+h)- f(x) - Df(x)(h) - \sum_{I \in \mathcal{I}_n^2} D_I f(x) h^I}{\|h\|^2} = 0$$

But $$h_1^{i_1}\cdots h_n^{i_n} = h^I < \|h\|^2 = h_1^2 + \cdots+ h_n^2$$ (by AM-GM inequality)

Thus we have: $$\lim_{h \rightarrow 0} \frac{\sum_{I \in \mathcal{I}_n^2} D_I f(x) h^I}{\|h\|^2} = 0$$

Then we are left with $$\lim_{h \rightarrow 0} \frac{f(x+h)- f(x) - Df(x)(h)}{\|h\|^2} = 0$$ showing $$|\frac{f(x+h)- f(x) - Df(x)(h)}{\|h\|^2}| < \varepsilon$$

Is this correct? I didn't use $$"A"$$ as a closed rectangle anywhere here.

• If the set is closed, does $f(x+h)$ make sense for every $x$ on the set? You need to be careful with that... Sep 13, 2021 at 0:34