Just like the title, find a Lipschitz constant for $f(t,x)=x(\sin(t)-x^2)$ in the region $0 \le t \le 2\pi$, $|x| \le M$
I know that I have $|f(x)-f(y)| \le L |x-y| $ where I need to kind of 'break down' my equation and then find an L based on my interval, but I've never seen an equation written with the $t$ in it as above. I've done a couple of these before, but none have depended on $t$ which is where I'm thrown off.
Computing the differential of $f,$ you get $$\nabla f(t,x)=(x\cos(t),\sin(t)-3x^2).$$ As we have $||\nabla f(t,x)||^2=x^2\cos^2(t)+(\sin(t)-3x^2)^2,$ so for $0\leq t\leq 2\pi$ and $|x|\leq M$ it gives $$||\nabla f(t,x)||^2\leq x^2+1+9x^4+6x^2\leq 1+ 7M^2+9M^4.$$ Finally, using the mean value inequality, I get for $t,t'\in[0,2\pi]$ and $x,x'\in[-M,M]$ : $$|f(t',x')-f(t,x)|\leq\sup\limits_{(t'',x'')\in[(t,x),(t',x')]}||\nabla f(t'',x'')||\cdot|(t',x')-(t,x)|\leq L\cdot||(t',x')-(t,x)||$$ with $L=1+ 7M^2+9M^4.$