I have to prove that a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable in $0$ knowing that $|f(x)| \leq \lVert{x}\rVert^2$. \ 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?

  • $\begingroup$ Use the definition of differentiability. This gives the result immediately. (*) is not true. $\endgroup$
    – Fnacool
    Mar 18 '16 at 18:28

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.

  • $\begingroup$ Thank you for answering. Can you explain a bit more why taking the limit of $\dfrac{|f(x)|}{\lVert{x}\rVert}$ shows that $f$ is differentiable? $\endgroup$
    – user302980
    Mar 18 '16 at 18:32
  • $\begingroup$ The derivative (when it exists) is a vector $Df(x)$ with the property that $$\lim_{y \to x} \frac{\lvert f(y) - f(x) - Df(x) \cdot (y-x) \rvert}{ \| y-x \|} = 0.$$ Here we have proven that the above limit is satisfied for our $f$ when $x = 0$ with $Df(x) = 0$ which means that the derivative exists at $x = 0$ and is zero there. $\endgroup$
    – User8128
    Mar 18 '16 at 21:56

It's easiest to consider this over the real line first, i.e. let $\mathrm f : \mathbb R \to \mathbb R$, where $|\mathrm f(x)| \le |x|^2$.

By definition, $\mathrm f$ is differentiable at $x=0$ if the following limit exists: $$L := \lim_{h \to 0} \left(\frac{\mathrm f(0+h)-\mathrm f(0)}{h}\right)$$

Since $|\mathrm f(x)| \le |x|^2$, we have $|\mathrm f(0)| \le |0|^2$, i.e. $\mathrm f(0)=0$. Hence $$L = \lim_{h \to 0} \left(\frac{\mathrm f(h)}{h}\right)$$

If $h > 0$ then $|h|^2=h^2$ and so: $$\lim_{h \to 0^+} \left(\frac{\mathrm f(h)}{h}\right) \le \lim_{h \to 0^+} \left(\frac{|h|^2}{h}\right)=\lim_{h \to 0^+} \left(\frac{h^2}{h}\right) = \lim_{h \to 0^+} \left(h\right) = 0$$

If $h < 0$ then $|h|^2=h^2$ and so: $$\lim_{h \to 0^-} \left(\frac{\mathrm f(h)}{h}\right) \le \lim_{h \to 0^-} \left(\frac{|h|^2}{h}\right)=\lim_{h \to 0^-} \left(\frac{h^2}{h}\right) = \lim_{h \to 0^-} \left(h\right) = 0$$

You can apply this argument to any continuous function $\mathrm f : \mathbb R^n \to \mathbb R$ be approaching zero along the different coordinate axes. For example, let $x=h$, $y=0$ and $z=0$. Since differentiation is linear, if you can prove dirrerentiability using a basis of directions then you're done.


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