I want to show this:

$$S=\sup_{s\in{[0,1]}}\int_0^1|(1-\cos(st^2))s| \, dt<1$$

  1. Since $1-\cos(st^2)$ is decreasing (fixed a $s$) for $t \in[0,1]$ so we have this: $$\sup_{s\in{[0,1]}}s\int_0^11-\cos(st^2) \, dt$$

  2. So if I prove $\int_0^1 1-\cos(st^2) \, dt< 1$ then we have $S<1$.

Setting $x=st^2$ $$1-\cos(st^2)=1-\cos(x) $$

Since $$\cos(x)>\frac{x}{2}\quad \forall x\in[0,1]$$

we have $$1-\cos(x)<1-\frac{x}{2}$$ then$$\int_0^11-\cos(st^2) dt<\int_0^1 1-\frac{s t^2}{2}dt= 1-\frac{s}{6}=\frac{6-s}{6}$$

so $$\sup_{s\in[0,1]} s\left(\frac{6-s}{6}\right)<1$$

Is this correct? Please let me know if some point is wrong, thanks1!


I could see one wrong, in point 1 of your solution, you said $1-cos(st^2)$ is decreasing on the interval for $s$ being constant, and $t\in [0,1]$ whereas the function should be increasing. If you think properly you surely can get that. In other parts I could not find any problem. If there is any retrospective effect of the mistake after considering the error I pointed out and then check out the places and see the solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.