On negation of lipschiz continuity

Question

Let $f: [a,b] \to R$ be continuous function which is not Lipschitz continuous.

Can we say there exist $x \in [a,b] $ and strictly monotone sequences, $\{x_n\}_{n=1}^{\infty} \subseteq [a,b] $ and $\lambda_{n} \in \mathbb{R^+} $ such that $x_n \to x$ and $\lambda_{n} \to + \infty $,

$$|f(x_n) - f(x)| > \lambda_{n} |x_n - x| $$

for all $n \in \mathbb{N}$?

P.S: Cleary we don't need to care about monotonicity of those two sequences!

Answer 1

Consider the function $$ f(x) = \begin{cases} x \sin (1/x), &\text{if}\ x\neq 0,\\ 0, & \text{if}\ x = 0. \end{cases} $$ This function is continuous in $\mathbb{R}$, it is not Lipschitz continuous, and it has continuous derivative in $\mathbb{R}\setminus\{0\}$, hence around each point $x\neq 0$ it is locally Lipschitz continuous.

Hence the only candidate for the point $x$ in your claim is $x=0$. On the other hand $$ |f(y) - f(0)| \leq |y| \qquad \forall y. $$

Answer 2

No, this is false. Consider a function defined on $[0,1]$ that mainly follows the line $y = 0$, but as $x \to 0$ has infinitely many increasingly steep peaks up to the line $y = x$.

(Formally, the function is zero, except on intervals $[1/n - 1/3n^2, 1/n + 1/3n^2]$, where it is piecewise linear, consisting of a triangle with vertices at the points $(1/n - 1/3n^2,0)$, $(1/n,1/n)$ and $(1/n + 1/3n^2,0)$.)

Answer 3

Here is a simplified version of @user49640's example: Consider a function $f : [0,1]\to\Bbb{R}$ defined by

$$ f(x) = \begin{cases} x \sin(1/x), & x > 0 \\ 0, & x = 0. \end{cases} $$

This function is pointwise Lipschiztz: For each $a \in [0, 1]$, there exists $\lambda_a \in [0,\infty)$ such that

$$\forall x \in [0, 1] : \qquad |f(x) - f(a)| \leq \lambda_a|x - a|. $$

On the other hand, $f$ is not Lipschitz: for $y_n = (2\pi n)^{-1}$ and $x_n = \left( 2\pi n + \frac{\pi}{2} \right)^{-1}$,

$$ \left| \frac{f(y_n) - f(x_n)}{y_n - x_n}\right| = \frac{x_n}{y_n - x_n} \to \infty \quad\text{as}\quad n\to\infty. $$

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p.s. Definitely this answer is preceded by another with exactly the same example in one minute. If the community thinks this is answer is redundant or repetitive I will gladly delete it.

Answer 4

Thanks for your answers ! I had a hard time to pick the best answer!

I've never herd about Pointwise Lipschitzian! (See @Sangchul Lee 's answer)

It turns out that any differentiable function $f:[a,b] \to \mathbb{R} $ is Pointwise Lipschitz! (super easy to prove ) So in order to reject my claim, we only need think about a (differentiable) function whose derivative is unbounded ! Now there are Tons of examples of such function, some typical ones are $x^{\alpha} \sin \frac{1}{x}$ where $ 1 < \alpha <2 $ ! This shows even I add the assumption of differentiability still my claim is not correct !

I am ashamed of asking such a dumb question :)