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I'm trying to prove/disprove the following statement:

If $|f(x)|\leq x^2$ for all $x$, then $f$ is differentiable at $x=0$.

My initial attempt: $|f(0)|\leq0\Rightarrow f(0)=0$. Then, because $-x^2\leq f(x)\leq x^2$ we get that $f'(0)=\lim\limits_{x\to0} \frac{f(x)-0}{x-0}=0$ (using Squeeze rule). But it appears to be wrong and I can't find a counter example. Any suggestions?

ikj
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  • Why does it appear to be false? – Bernard Jul 16 '15 at 17:55
  • It appears to be correct. (In fact, it is correct.) – David Mitra Jul 16 '15 at 17:55
  • Why do you think this is wrong? It's fine, more or less. Now, you left out some details - could be that when you wrote down the details of the application of that Squeeze thing there was a problem. – David C. Ullrich Jul 16 '15 at 17:56
  • Write it like this: $-|f(x)/x|\le f(x)/x \le| f(x)/x|$. Does this clear things up? – David Mitra Jul 16 '15 at 17:59
  • @trfv And false for $x<0$. This is exactly the detail I was conjecturing may have been what someone said was wrong. If you say $-x^2\le f\le x^2$, hence $-x\le f(x)/x \le x$ then yes that is wrong when $x<0$. Easy to fix: Just say $0\le |f(x)|\le x^2$< hence for $x\ne0$ we have $0\le|f(x)/x|\le|x|$, and squeeze that. – David C. Ullrich Jul 16 '15 at 17:59
  • @trfv The floor function is not a counterexample. Look at $x=-0.1$. – David K Jul 16 '15 at 18:01
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    The function $x^2\sin(1/x^3)$ is differentiable everywhere, but the derivative is not continuous at $0$. – Empy2 Jul 16 '15 at 18:02
  • @Michael The derivative doesn't have to be continuous for its existence. The example you discussed shows this. – Mark Viola Jul 16 '15 at 18:06
  • @trfv No worry. ;-)) – Mark Viola Jul 16 '15 at 18:09
  • @Dr.MV - actually, Michael's example works. $|x^2\sin(1/x^3)|\leq x^2$ is indeed true for all $x$ but the derivative of that function is undefined at $x=0$. – ikj Jul 16 '15 at 18:14
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    @user254036 That is incorrect. The derivative at $0$ exists and is given by $$\lim_{h\to 0}\frac{h^2\sin(1/h^3)-0}{h}=0$$But, the derivative is discontinuous since for $x\ne0$, the limit as $x\to 0$ of the derived function does not exist. Two very different things. – Mark Viola Jul 16 '15 at 18:22
  • @Dr.MV - thanks! By the way - now I see. His example is incorrect at least because the function is not defined at $x=0$. – ikj Jul 16 '15 at 19:01

2 Answers2

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$$\left|\frac{f(x)-f(0)}{x}\right|\le\left|\frac{x^2-0}{x}\right|=|x|\to 0$$

as $x\to 0$ The limit of the left-hand side is $f'(0)=0$.

Mark Viola
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Dr. MV's solution is correct; Here, I am only willing to comment a few occasions where we could/couldn't prove the result.

Remark 1: If there existed a $\lambda > 1$ such that $|f(x)|\le |x|^{\lambda}$, the result would follow by the same reasoning. This has to do with the Hölder Cotinuity exponent of the function at $x=0$.

Def.: We call $f : \Bbb{R} \rightarrow \Bbb{R} $Hölder continuous of exponent $\lambda > 0$ at the point $x_0$ if we can assure that the inequality $|f(y)-f(x_0)|\le C |y-x_0|^{\lambda}$ for all $y$. ($C$ is a fixed constant)

We then call a function *Hölder continuous of exponent*$\lambda$ if the above condition holds at every $x_0$.

It can be (easily) showed that the only Hölder continuous functions of exponent $\lambda > 1$ are the constant functions. For $\lambda = 1$, we recover the concept of Lipschitz functions - that are far from being 'trivial' in the former sense, but which are 'essentially' differentiable functions -, and for $0<\lambda <1$ we have a wide class of non-differentiable functions - for example, a wide class of irregular continuous functions, such as the Brownian Motion's paths, are Hölder Continuous of some exponent.

Remark 2: The previous remark (maybe) suggests us that the whole thing must fail even for $\lambda = 1$. For example, $f(x)=|x|$ is not even differentiable at $x=0$.

For example, the Brownian Motion's paths are locally Hölder Coninuous of exponent $\forall \; \alpha <1/2$, but they are nowhere differentiable almost surely. This provides some more examples of how can a function be relatively well-behaved but not differentiable.

João Ramos
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  • You wrote "It can be (easily) showed that the only Hölder continuous functions of exponent λ>1 are the constant functions" Please clarify. If we have an analytic function with zero derivative at a point, but say non-zero second derivative, then surely that function is Holder continuous with $\lambda=2$ at that point. – Mark Viola Jul 17 '15 at 03:38
  • Well, I meant to say Globally Hölder. It is clear on the post that there is a distinction between being Hölder continuous at a point and globally Hölder (which I called only Hölder continuous) – João Ramos Jul 17 '15 at 13:03