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This previous post of mine was closed by people who were saying that these two limits always evaluate to the same thing:Can these two limits be used to directly check the twice differentiability of a function?. Here, I'm giving an example, to show that these two limits don't always evaluate to the same thing and hence can be used as a second order differentiability check.

First, consider this function: $$f(x)=\frac{x^2}{2},x\geq0$$ $$=\frac{-x^2}{2},x<0$$, when $x$ is greater than or equal to 0

Now, clearly this function is once differentiable at $x=0$. But after the first differentiation, the derivative of this function comes out to be $|x|$, which is certainly not differentiable at $x=0$. So, this function is not twice differentiable at $x=0$.

Now these are the limits I'm using to check twice differentiability: $$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ And,$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$ When these limits are applied to $f(x)$ at $x=0$, the first limit evaluates to 1 while the second limit evaluates to -1. So, you, see these two limits are not always equal. And, hence the given function is not twice differentiable at $x=0$.

Now, please tell me if these two limits can be used as a second-order differentiability check or not.

Here are my calculations:

For the function: $$f(x)=\frac{x^2}{2},x\geq0$$ $$=\frac{-x^2}{2},x<0$$. Now, this function is differentiable once at $x=0$. But after the first differentiation, its derivative comes out to be $|x|$ which isn't differentiable at $x=0$. So, this function is not twice differentiable at $x=0$. So, those limits should give different values. Now,$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ At $x=0$, $$=\lim_{h\rightarrow 0}\frac{f(2h)-2f(h)+f(0)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{\frac{(2h)^2}{2}-2\frac{h^2}{2}+0}{h^2}$$ $$=2-1=1$$ And, at $x=0$, $$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{f(-2h)-2f(-h)+f(0)}{h^2}$$ $$=\lim_{h\rightarrow 0}\frac{\frac{-(-2h)^2}{2}+2\frac{-h^2}{2}+0}{h^2}$$ $$=-2+1=-1$$ So, these limits are not equal, which means $f(x)$ is not twice differentiable at $x=0$. So, I think this limit works. Is there some way to prove that a function is twice differentiable only if these two limits are equal?

Please note that $h$ itself is assumed to be positive.

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    There is no question here. – 2017-02-28
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    @Siminore See the last line. – 2017-02-28
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    Do you have a definition of differentiability check? If so, can you test your idea against this definition? – 2017-02-28
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    Anyway, the two limits are necessarily equal, if they exist. Please remember that $+2h$ does not imply $h>0$. – 2017-02-28
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    The limit doesn't exist here. The _one-sided limits_ exist and are different, but the limit doesn't exist. – 2017-02-28
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    @Siminore What else does it imply? – 2017-02-28
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    Any symbol like $-h$ does not necessarily represent a negative number. It may be positive if $h$ itself is negative. You can not attribute sign to a symbol, you have to mention it explicitly that the symbol denotes a positive or a negative number as the case may be. In the absence of such information the symbol may represent any real number. – 2017-02-28
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    @ParamanandSingh $h$ is positive. Now, is this correct? – 2017-02-28
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    If $h$ is positive then you better replace $h\to 0$ with $h\to 0^{+}$. This is the proper way to state that $h$ is positive and tends to $0$. – 2017-02-28
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    @ParamanandSingh You're right. But usually $h\rightarrow 0^+$ or $h\rightarrow 0^-$ is not used because $h$ is already assumed to be positive. Because remember that $h$ is actually obtained by a change of variable in the original definition of the derivative at $x=c$ which is $\lim_{x\rightarrow c^+}\frac{f(x)-f(c)}{x-c}$. So, it's already positive. – 2017-02-28
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    I think this is where you are mistaken. You must explicitly state the sign of $h$ by using notation like $h\to 0^{+}$ otherwise $h$ can have both positive and negative values. – 2017-02-28

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