Use Mean Value value theorem to check whether the following functions are Lipschitz continuous

Question

Use Mean Value value theorem to check whether the following functions are Lipschitz continuous :

(i) $ f(x)=x+\frac{1}{x} , \ \ on \ \ (0, \infty) \\ (ii) \ \ f(t,x)=\frac {\sin (tx)}{x} \ \ \ on \ \ (0,\infty) $

Answer:

(i)

$ f(x)=x+\frac{1}{x} \ $

Now $ \ \ f'(x)=1-\frac{1}{x^2} \ $ is unbounded near $ \ x=0 \ $

So $ \ \ |f(x_1)-f(x_2) | \nless \ |f'(x)| |x_1-x_2| \ $ near $ \ x=0 \ $

I think $ f(x)=x+\frac{1}{x} \ $ is Lipschitz in $ \ [1,\infty) \ $

Am I right ?

(ii)

$ f(t,x)=\frac{\sin (tx)}{x} \ , \ t \in (-\infty, \infty) , \ x \in (-\infty, \infty) $

Now,

$ \frac{\partial f}{\partial x}=\frac{t \cos (tx)}{x}-\frac{\sin (tx)}{x^2} \ $

Thus,

$ | \frac{\partial f}{\partial x} | \leq |\frac{t}{x} | +|\frac{1}{x^2} | \ $ can be unbounded when $ t \to \infty \ \ \ and \ \ \ x \to 0 \ $

Thus , $ f(t,x)=\frac{\sin (tx)}{x} \ $ is not Lipschitz in $ \ (-\infty, \infty) \ $

Am I right ?

Answer 1

using the mean value Theorem we have $$\frac{f(b)-f(a)}{b-a}=f'(\xi)=1-\frac{1}{\xi^2}$$ this is not bounded on the given interval for your second function we have $$\frac{f(t,b)-f(t,a)}{b-a}=\frac{t\xi\cos(t\xi)-\sin(t\xi)}{\xi^2}$$ this is also not bounded on the given interval

Answer 2

The first one is not Lipschitz continuous on the interval $(0,\infty)$, but it is Lipschitz continuous on any interval $[a,\infty)$ for $a>0$, in which case the Lipschitz constant can be taken as $1+1/a^{2}$.

Why it is not Lipschitz on $(0,\infty)$? Assume the contrary it were, then some constant $M>0$ is such that $|f(x)-f(y)|/|x-y|\leq M$ for any $x\ne y$, $x,y\in(0,\infty)$. Taking $y\rightarrow x$, the left side is exactly $|f'(x)|=|1-x^{-2}|$, so $|1-x^{-2}|\leq M$ for any $x\in(0,\infty)$. Taking $x\rightarrow 0^{+}$ the expression will blow up, a contradiction.

Answer 3

For the second example, we may suppose that $t>0$ (and $t$ is fixed), we have $\displaystyle f(t,x)=\int_0^t \cos(ux)du$, hence as $|\cos(y_1)-\cos(y_2)|\leq |y_1-y_2|$ for all $y_1,y_2$, we get that $$|f(t,b)-f(t,a)|\leq \int_0^t|\cos(ub)-\cos(ua)|du\leq \frac{t^2}{2}|b-a|$$ and $f(t,x)$ is Lipschitz-continuous.(And of course, this imply that the derivative of $f(t,x)$ with respect to $x$ is bounded on $\mathbb{R}$, by $t^2/2$).