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The function $F(x)$ is said to be smooth at the point $x$, if $$\lim_{h\to 0} \frac{F(x+h)+F(x-h)-2F(x)}{h}=0.$$ My question is that if function $F$ is differentiable at some point then how can we show that function $F$ is smooth?

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    Did you possibly forget the square in the denominator?2012-06-05
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    No it is actually $h$ in the denominator of the definition of smooth function.2012-06-05
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    @Gregor: yes, it is suppose to be $h$. The expression is saying that "the second order difference" is $o(h)$.2012-06-05
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    Your edit, on the other hand, is wrong. If $F'(x)$ exists, you expect $$\lim_{h\to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h\to 0} \frac{F(x) - F(x-h)}{h}$$ since the left and right limits should be equal, so the limit of their difference should be 0, not $F'(x)$.2012-06-05

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For once this is the weirdest definition of smooth I have seen so far, but working with the definition just observe $$\frac{F(x+h)+F(x-h)-2F(x)}{h}=\frac{F(x+h)-F(x)}{h}+\frac{F(x-h)-F(x)}{h}$$

If $F$ is differentiable the first summand converges to $F'(x)$ and for the second summand we get by substituting $h\mapsto -h$ $$-\frac{F(x+h)-F(x)}{h} .$$ Taking the limit gives the desired result. We may change the sign of $h$ above since $h\to 0\Leftrightarrow -h\to 0$.

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    Is converse true? I think converse is not true by taking the example $f(x)=|x|$. What you say about this?2012-06-05
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    Well, it is not the usual definition of smooth for sure. That symmetric second order differences is 0 does not actually guarantee that $F'$ exists. Take $F(x) = +\sqrt{x}$ for $x \geq 0$ and $F(x) = -\sqrt{|x|}$ for $x < 0$. The expression $F(x+h) + F(x-h) - 2 F(x)$ evaluates identically to 0 when $x = 0$. But the function is not differentiable in the usual calculus sense at 0 (since the tangent is vertical). In fact, you can take any function $G$ which is non-differentiable at 0, and set $F(x) = G(x)$ for $x > 0$ and $F(x) = -G(-x)$ for $x < 0$ and $F(0) = 0$ to get more examples.2012-06-05
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    $f(x) = |x|$ is _not_ a counter example. At $0$, your expression evaluates to $\pm 2$ (depending if you take the left or right limits). So it is not smooth by the definition.2012-06-05
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    $f(x)=-1$ for $x<0$ and $f(x)=+1$ for $x>0$ and $f(0)=0$ is a counterexample. So $f$ doesn't even need to be continuous.2012-06-05
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    Another example which is as simple as @ThomasAndrews' but a bit more genreal: Take any differentiable function $f$ and consider $F(y)=f(y)$ for $y\leq x$ and $F(y)=f(y)+c$ for $y>x$ and $c\neq 0$.2012-06-05
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    If $f(x)$ is any odd continuous function ($f(-x)=-f(x)$ for all $x$, and thus also $f(0)=0$) then $f(x)$ is smooth at $x=0$, because $f(h)+f(-h)-2f(0)=0$.2012-06-05
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    @SimonMarkett You still need $F(x)=f(x)+\frac{c}{2}$ to show that it is smooth at $x$2012-06-05
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    @Thomas: Fair enough. Thanks2012-06-05