The function $f(x)=x\log(x)$ is clearly continuous on $[0,1]$ and thus uniformly continuous.

I would like to ask whether one can explicitly state a $\delta(\varepsilon)$ such that for all $x,y \in [0,1]$

$$\left\lvert x-y \right\rvert \le \delta \Rightarrow \left\lvert f(x)-f(y) \right\rvert< \varepsilon?$$

  • $\begingroup$ Try replicating a general proof that continuity on a compact interval implies uniform continuity on that interval, using a proof that is constructive in the sense that uniform continuity is proven by finding an explicit $\delta$. $\endgroup$ – MathematicsStudent1122 Aug 12 '18 at 23:23
  • $\begingroup$ For instance, the second proof in the first answer here would work. Note that since pointwise continuity is assumed, the proof uses $\delta_{x_0}(\epsilon)$'s which demonstrate pointwise continuity at $x_0$. You can find explicit pointwise-continuity $\delta_{x_0}(\epsilon)$s for $x \log x$ by replicating a proof that the product of continuous functions is continuous. $\endgroup$ – MathematicsStudent1122 Aug 12 '18 at 23:23
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    $\begingroup$ but log(x) is not continuous at zero. $\endgroup$ – Sascha Aug 12 '18 at 23:49
  • $\begingroup$ Not an issue, just deal with $x_0 = 0$ as a special case. $\endgroup$ – MathematicsStudent1122 Aug 13 '18 at 0:32

For the direct $\epsilon-\delta$ approach, assuming WLOG $x > y > 0$ we have

$$|x \log x - y \log y| \leqslant |x-y| |\log x| + |y| |\log x - \log y|\\ = |\sqrt{x}- \sqrt{y}||\sqrt{x}+\sqrt{y}||2 \log \sqrt{x}| + |y| |\log \left(1 + \frac{x-y}{y} \right)$$

Applying the inequalities $|\log s| < (1-s)/s$ for $s \in (0,1]$ and $\log(1+t) < t$ for $t > 0$ with $s = \sqrt{x}$ and $t = (x-y)/y$ we get

$$\tag{*}|x \log x - y \log y| \leqslant \frac{2|\sqrt{x} + \sqrt{y}|}{\sqrt{x}}(1- \sqrt{x})|\sqrt{x} - \sqrt{y}| + |y|\frac{|x-y|}{|y|}\\ \frac{4\sqrt{x}}{\sqrt{x}}(1- \sqrt{x})|\sqrt{x} - \sqrt{y}| + |y|\frac{|x-y|}{|y|}\\ \leqslant 4|\sqrt{x} - \sqrt{y}| + |x-y|$$

Since, $x \mapsto \sqrt{x}$ and $x \mapsto x$ are uniformly continuous on $[0,1]$, we can find $\delta(\epsilon) > 0$ such that the RHS of (*) is less than $\epsilon$ when $|x-y| < \delta(\epsilon)$ for all $x,y \in (0,1]$.

If the function $f$ is defined such that $f(0)= 0$, then this can be handled using $\lim_{x \to 0} z \log z = 0$, extending uniform continuity to $[0,1]$.


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