On negation of lipschiz continuity 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! 
 A: 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
$$
|f(y) - f(0)| \leq |y| \qquad \forall y.
$$
A: 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 $(1/n + 1/3n^2,0)$.)
A: 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 $x_n = \left( 2\pi n + \frac{\pi}{2} \right)^{-1}$, 
$$ \left| \frac{f(y_n) - f(x_n)}{y_n - x_n}\right| = \frac{x_n}{y_n - x_n} \to \infty \quad\text{as}\quad n\to\infty. $$

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.
A: 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 :)
