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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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    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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    @Thomas: Fair enough. Thanks2012-06-05