Find $\delta(\varepsilon)$ function

Question

Find such $\delta(\varepsilon): \mathbb{R_+} \rightarrow \mathbb{R_+}$ that for $f: [0, 1] \rightarrow \mathbb{R}$ and $f=\begin{cases}x\log x\text{, if x $\ne$ 0}\\0\text{, if x = 0}\end{cases}$ $|x-y|\le\delta(\varepsilon) \Rightarrow |f(x) - f(y)|\le \varepsilon $. I don't know how to bound $|x\log x - y\log y| \le \varepsilon$. Please, give me a hint

Answer 1

You don't actually have to bound $|x\log x - y\log y|$

You can prove that $f$ is continuous in $[0,1]$ because $lim_{x\rightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem $$\exists M>0, \forall(x,y)\in ]0,1[,\ \ M|x-y| < |f(x)-f(y)|$$ You can then chose $\delta(\epsilon) = M = \sup_{x\in[0,1]}|f'(x)| = constant$ because you proved that $f'$ is continuous