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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.

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

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