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.

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

1 Answer 1


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


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