- #1
michonamona
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Homework Statement
Let r>0, and let f be a function from [tex]B_{r}(\textbf{0}) \rightarrow \textbf{R} [/tex], and suppose that there exists an [tex]\alpha > 1[/tex] such that [tex]|f(\textbf{x})| \leq ||\textbf{x}||^{\alpha} [/tex] for all [tex]\textbf{x} \in B_{r}(\textbf{0})[/tex]. Prove that f is differentiable at 0.
What happens to this result when [tex] \alpha = 1[/tex]?
Homework Equations
Note:
The characters in bold are vectors.
[tex]B_{r}(\textbf{0})[/tex] is an open ball in [tex]R^{n}[/tex] about 0
The symbol ||.|| represents the norm of the vector.
Some definitions of differentiability of a function
A function f from [tex]R \rightarrow R^{n} [/tex] is differentiable if
[tex]lim_{\textbf{h} \rightarrow \textbf{0}} \frac{f(\textbf{0} + \textbf{h}) - f(\textbf{0}) - Df(\textbf{a})(\textbf{h})}{||\textbf{h}||} = \textbf{0} [/tex].
Where [tex]Df(\textbf{a})[/tex] is the matrix of first order partials. In our case, since the function is from R to R^n, this would be the gradient of f.
The Attempt at a Solution
The started this problem by computing that the first order partials exists. But that by itself is not sufficient to concluded that f is differentiable at 0. The next step would be to show that the first-order partials of f are continuous at 0. This is where I got stuck.
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