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Let $f$ be a function on $[0,1]$ into $Bbb{R}$. Suppose that if $xin[0,1],$ there exists $K_x$ such that begin{align}|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].end{align}
Prove or disprove that there exists $K$ such that begin{align}|f(x)-f(y)|leq K |x-y|,;forall;;x,yin[0,1].end{align}

DISPROOF

Consider the function begin{align} f:[0&,1]to
Bbb{R}, \&xmapsto sqrt{x} end{align}

Let $x=0$ and $yin (0,1]$ be fixed. Then,
begin{align} left| f(0)-f(y) right|&=left|0-sqrt{y} right| end{align}
Take $y=1/(4n^2)$ for all $n.$ Then,
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=left|0-dfrac{1}{sqrt{4n^2}} right| \&=2n^{3/2}left|dfrac{1}{4n^2} -0 right| end{align}
By assumption, there exists $K_0$ such that
begin{align} left| f(0)-fleft(dfrac{1}{4n^2}right) right|&=2n^{3/2}left|dfrac{1}{4n^2} -0 right|leq K_0left|dfrac{1}{4n^2} -0 right| end{align}
Sending $ntoinfty,$ we have
begin{align} infty leq K_0<infty,;;text{contradiction}. end{align}
Hence, the function $f$ is not Lipschitz in $[0,1]$.

QUESTION: Is my disproof correct?

You are correct, the function $f$ is not Lipschitz in $[0,1]$, but your argument should be modified. You may simply say that
$$frac{f(1/n)-f(0)}{frac{1}{n}-0}=sqrt{n}to +infty$$
which contradicts the fact that $|f(x)-f(y)|/|x-y|$ is bounded by a constant $K$.

On the other hand this is not a counterexample for your statement. For the same reason as above (Just take $y=1/n$), for $x=0$ there is no constant $K_0$ such that
$$|f(0)-f(y)|leq K_0 |0-y|,;;forall yin[0,1].$$
Instead consider the function
$$f(x)=cases{xsin(1/x)& if $xnot=0$\0& if $x=0$,}$$
If $x_n=1/(2pi n)$ and $y_n=1/(2pi n+pi/2)$.
then $|f(x_n)-f(y_n)|/|x_n-y_n|$ is unbounded which implies that $f$ is not Lipschitz in $[0,1]$, but for any $xin[0,1],$ there exists $K_x$ such that $$|f(x)-f(y)|leq K_x |x-y|,;;forall;;yin[0,1].$$
In fact take $K_0=1$ and for $xin(0,1]$ the existence of $K_x$ follows from
$f'in C^1((0,1])$.

This isn't incorrect per se, but its unnecessarily convoluted. To show that $f'$ is unbounded, just observe that $f'(x)=frac{1}{2sqrt{x}}$ so that $lim_{xto 0^+}f'(x)=infty$. Hence, $f$ is not Lipschitz on $[0,1]$, which is precisely what you want to prove. That's all you need to say.

The last part is a little hand-wavy though. Technically, you should say something like this: $f$ is Lipschitz on $[epsilon,1]$ with constant $K_epsilon:=sup_{epsilonleq xleq 1}|f'(x)|$. Since $K_epsilontoinfty$ as $epsilonto 0^+$, $f$ is not Lipschitz on $[0,1]$.

Better still just to do an explicit calculation, as the other fellow did.