Consider the function $f(x)=\cos(x)-\cos((1+t)x)$ for $t>0$ on $x\in\mathbb{R}$. I would like to show the following claim which is obvious from looking at the graph of the function.
Claim. If $0<t<1$ and $\varepsilon>0$ then there is $x\in\mathbb{R}$ with $|f(x)|>2-\varepsilon$.
What I tried: First I tried to characterise the local maxima, by looking at $f'(x)=(1+t)\sin((1+t)x)-\sin(x)$. However the equation we obtain by setting $f'(x)=0$ is probably not solvable. Or does somebody know a trick to solve $$ \frac{\sin(x)}{\sin((1+t)x)}=1+t? $$ Second I tried to find an $x\in\mathbb{R}$ that is close to $2k\pi$ and for which $(1+t)x$ is close to $(2l+1)\pi$ for $k,l\in\mathbb{N}$. However when I do this I run into an issue I cannot solve. For example: Set $x=\frac{(2l+1)\pi}{1+t}$ for $l$ to be determined. Since the rationals are dense in the continuum we can certainly find $m,n\in\mathbb{R}$ so that $\frac{m}{n}$ is close to $\frac{1}{1+t}$. Perhaps we can also choose $m=2k$ to be even and $n=2l+1$ to be odd. Hence for all $\varepsilon>0$ we have $$ x=\frac{n\pi}{1+t}=(\frac{1}{1+t}-\frac{m}{n})n\pi+m\pi\leq\varepsilon n\pi+m\pi. $$ It seems that no matter what I do, the $\varepsilon$ will be in a product with $n$, however $n$ depends on $\varepsilon$ and thus my proof does not show that $x$ is close to $m\pi$. How to get rid of this problem?
Context. The goal is to show that the semigroup $T_tu(x)=u((1+t)x)$ for $x\in\mathbb{R},t\geq0,u\in C_b(\mathbb{R})$ is not strongly continuous. Obviously any periodic function will do the job, so if you suggest I should better look at a simpler version of $cos$ like the sharktooth-function, then let me know. Edit: So something to note is that these operators do not form a semigroup. In order to solve it, take $T_tu(x)=u(e^t x)$ instead. By continuity of the exponential the calculation will still work.