I have to prove that a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable in $0$ knowing that $|f(x)| \leq \lVert{x}\rVert^2$.
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This is what I have:
$ 0 \leq |f(0)| \leq\lVert{0}\rVert^2 = 0 \Rightarrow f(0) = 0 $
We calculate the partial derivatives of $f$ in $0$:
\begin{eqnarray*} & & \forall v \in \mathbb{R}^n:Df|_{0}\cdot v = \lim_{t \rightarrow 0}\dfrac{f(0+tv)-f(0)}{t}\\ \\ & &-\lim_{t \rightarrow 0}\dfrac{\lVert{tv}\rVert ^2}{t} \leq \lim_{t \rightarrow
0}\dfrac{f(0+tv)-f(0)}{t} \leq \lim_{t \rightarrow
0}\dfrac{\lVert{tv}\rVert ^2}{t} \\ & & 0 = -\lim_{t \rightarrow
0}t\lVert{v}\rVert ^2 \leq \lim_{t \rightarrow
0}\dfrac{f(0+tv)-f(0)}{t} \leq \lim_{t \rightarrow 0}t\lVert{v}\rVert^2 = 0\\ \\ & & \Rightarrow Df|_{0}\cdot v = 0 \end{eqnarray*}In the point $0$ is every direction derivative $0$; which means the
partial derivatives are $0$ in $0$. Now we have to show that these
partial derivatives are continuous in $0$.We know that $|f(x)| \leq \lVert{x}\rVert^2_E$
$\Rightarrow |f(x_1,\ldots ,x_n)| \leq x_1^2+ \ldots +x_n^2$
$\Rightarrow |\dfrac{\partial f}{\partial x_i}(x_1, \ldots , x_n)|
\leq 2x_i$ (*)$ \forall i \in \{1,\ldots,n \},\forall \epsilon > 0, \exists \delta
:= \dfrac{\epsilon}{2}: \lVert{x-0}\rVert_E < \delta \Rightarrow
|\frac{\partial f}{\partial x_i}(x)-\frac{\partial f}{\partial x_i}(0)|\leq |2x_i – 0| \leq 2\rVert{x}\lVert_E \ < \ 2 \delta = \epsilon $This means that all the partial derivatives are continous, which means
the function is differentiable.
But I'm not sure about a lot of things that I wrote. For example is this step (*) correct?
Best Answer
I think your reasoning is basically correct. I think it is a bit less complicated than you are making it. We see $$\frac{\lvert f(x) \rvert}{\| x \|} \le \| x\|$$ for all $x \neq 0$. Taking the limit as $x \to 0$ shows that $f$ is differentiable at $0$ with $Df = 0$ there.