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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!

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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.

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