# For $f:\mathbb{R}^{n} \to \mathbb{R}$, for which $\alpha > 0$ does the condition $|f(x)| \leq |x|^{\alpha}$ imply $f$ is diff

For $$f:\mathbb{R}^{n} \to \mathbb{R}$$, for which $$\alpha > 0$$ does the condition $$|f(x)| \leq |x|^{\alpha}$$ imply $$f$$ is differentiable at $$0$$ and why?

I have seen a solution to this question and it is similar to mine, but I want to understand what was faulty with my reasoning compared to the correct solution.

My Attempt:

We are trying to find under what conditions $$\alpha > 0$$ would imply my function is differentiable at $$0$$. TO be differentiable at $$0$$ means there exists a value $$Df(0)$$ such that:

$$\lim_{h \to 0} \frac{f(0 + h) - f(0) - D(f(0) \cdot h}{|h|}\ \text{where}\ h = |x-a|$$

Observe: $$|f(0)| \leq |0|^{\alpha}$$

As such we break things up into cases:

Case 1: if $$f(0) = 0$$. This means

$$\lim_{h \to 0} \frac{|f(h)|}{|h|} \leq \frac{|h|^{\alpha}}{|h|} \\ \Rightarrow\ \lim_{h \to 0} |h|^{\alpha - 1} = 0$$

So we can conclude that when $$f(0) = 0$$ that as long as $$\alpha > 1$$, then our function will be differentiable.

Case 2: $$f(0) \neq 0$$.[This is where my approach falls apart]. Again we have the condition:

$$\lim_{h \to 0} \frac{f(0 + h) - f(0) - D(f(0) \cdot h}{|h|}$$

But my concern is that since we have the observation that $$|f(0)| \leq |0|^{\alpha}$$ then

$$\lim_{h \to 0} \frac{|f(0 + h) - f(0) - D(f(0) \cdot h|}{|h|} \geq \frac{|f(h)|}{|h|}$$

My version of Case 1, is pretty much in line with what the correct solution said, except for considering $$f(0) = 0$$. Should I not have looked at it in terms of $$f(0) = 0$$?

How could I reconcile this?

Edit: I think I just realized that the issue is that I overlooked the idea that I am considering the Norm of the function, which means that whatever concerns I had were actually captured in my Case 1. Let me know if this is the right reasoning.

• @Masacroso: The case $\alpha=1$ doesn't guarantee differentiability. Just take $f:\Bbb R\to\Bbb R$ given by $f(x)=|x|$. Dec 12, 2019 at 21:55

There is no need to break this into cases. If $$\alpha > 0$$ and $$|f(x)|\le |x|^\alpha,$$ then $$f(0)$$ must be $$0.$$