Is there a bounded function ff with f′f’ unbounded and f″f” bounded?

Is there a C2-function f:RR 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:RR
such that

+12πte(yx)2/2tf(y)dyf(x)=12π+ey2/2(f(x+ty)f(x))dy

does NOT uniformly converge to 0 for t0+. 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 yxn one has
|f(y)f(xn)|<M|yxn|
So if y is such that |yxn|<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

Leave a Comment