$f(t,x)=t^2\sin(x)\cos(x)$, S = {(t,x) : |t| ≤ a, |x| < ∞} for a positive constant a.
Hint: When evaluating $|f (t, x)−f (t, y)|$, add and subtract the term $\sin(x) \cos(y).$
I tried using the hint but I didn't know what to do without using some esoteric trig identity:
$f(t, x)−f(t, y)\\
=t^2sin(x)cos(x)-t^2sin(y)cos(y)\\
=t^2sin(x)cos(x)-t^2sin(y)cos(y)+sin(x)cos(y)-sin(x)cos(y)\\
=t^2sin(x)cos(x)+sin(x)cos(y)-t^2sin(y)cos(y)-sin(x)cos(y)\\
=sin(x)[t^2cos(x)+cos(y)]-cos(y)[t^2sin(y)-sin(x)]
$
So I then tried using the MVT since both $f(t,x)$ and $\frac{∂f}{∂x}$ are continuous and bounded on the interval, but I'm not sure about my work.
$\frac{f(t,x)-f(t,y)}{x-y}=\frac{∂f}{∂x}(c)\\
\frac{f(t,x)-f(t,y)}{x-y}=t^2cos^2(c)-t^2sin^2(c)\\
|\frac{f(t,x)-f(t,y)}{x-y}|=|t^2cos^2(c)-t^2sin^2(c)|\\
|\frac{f(t,x)-f(t,y)}{x-y}|=t^2|cos^2(c)-sin^2(c)|\\
|f(t,x)-f(t,y)|=t^2|cos^2(c)-sin^2(c)||x-y|\\
|f(t,x)-f(t,y)|≤t^2(1)|x-y|\\
|f(t,x)-f(t,y)|≤a^2|x-y|\\$
So $K=a^2$ and $f(t,x)$ is Lipschitz… Right?
Best Answer
I might be inclined to replace $\sin x\cos x$ with $\frac 12 \sin 2x$
$f(t,x) - f(t,y) = \frac 12t^2(\sin 2x - \sin 2y)$
$\sin (A+B) - \sin (A-B) = 2\cos A\sin B$
$\sin A - \sin B = \sin(\frac {A+B}2 + \frac {A-B}2) - \sin(\frac {A+B}2 - \frac {A-B}2)$
$\sin 2x - \sin 2y = 2\cos(x+y)\sin(x-y)$
$f(t,x) - f(t,y) = t^2(\cos(x+y)\sin(x-y))$
$|f(t,x) - f(t,y)|\le t^2\le \alpha^2$