Is there a C2-function f:R→R that is bounded and such that f′(x) is unbounded, but f″(x) is bounded again?

For example, f(x)=sin(x2) is bounded and has unbounded derivative f′(x), but its second derivative is also unbounded.edit:

Thanks for the great answer.

The reason I came up with this question, was the following:I’d like to find a bounded continuous function f:R→R

such that∫+∞−∞1√2πte−(y−x)2/2tf(y)dy−f(x)=1√2π∫+∞−∞e−y2/2(f(x+√ty)−f(x))dy

does NOT uniformly converge to 0 for t→0+. Any help would be much appreciated.

**Answer**

No, this can’t occur. Suppose f′(x) were unbounded but |f″(x)|<M for some M. Then for any N you could find some xn with |f′(xn)|>N. By the mean value theorem for

any y≠xn one has

|f′(y)−f′(xn)|<M|y−xn|

So if y is such that |y−xn|<N2M then

|f′(y)−f′(xn)|<MN2M=N2

Since |f′(xn)|>N, this would mean |f′(y)|>N2, and furthermore by continuity of f′, one necessarily has that f′(y) has the same sign as f′(xn). So integrating one has

|f(xn+N2M)−f(xn)|=|∫xn+N2Mxnf′(y)dy|

>|∫xn+N2MxnN2dy|=N24M

By the triangle inequality, |f(xn+N2M)−f(xn)|≤|f(xn+N2M)|+|f(xn)|. So by the above equation, at least one of |f(xn+N2M)| and |f(xn)| is greater than N28M. You can do this for any N, so f(x) must be unbounded.

**Attribution***Source : Link , Question Author : Steven , Answer Author : Mario Carneiro*