Does $f_n(x)=\frac{x}{n}\log(\frac{x}{n})$ converge uniformly in $(0,1)$?

Question

$f_n(x)=\frac{x}{n}\log\frac{x}{n}$

I tried to expand the bounds to $[0,1]$ and prove that the hypotheses of Dini's uniform convergence criterion are satisfied, but I'm not even sure I can expand the bounds since $f_n(0)$ isn't defined.

Answer 1

Use the inequality $-\log x < (1-x)/x$ for $0 < x < 1$.

Then

$$\left|\frac{x}{n} \log \frac{x}{n}\right|= -\frac{x}{n} \log \frac{x}{n}=-2\frac{x}{n} \log \frac{\sqrt{x}}{\sqrt{n}}<\frac{2}{\sqrt{n}}\sqrt{x}\left(1-\frac{\sqrt{x}}{\sqrt{n}}\right)<\frac{2}{\sqrt{n}},$$

and

$$\sup_{x \in (0,1)}\left|\frac{x}{n} \log \frac{x}{n}\right|< \frac{2}{\sqrt{n}} \to 0$$

Answer 2

Hint: $$ f_n(x)=\frac{x\log x}{n}-\frac{\log n}{n}\,x $$ and $\lim_{x\to0^+}x\log x=0$.

Answer 3

First, let us compute the limit of this sequence of functions:

$$\lim \limits _{n \to \infty} \frac x n \log \frac x n = \lim \limits _{t \to 0, \ t > 0} t \log t = \lim \limits _{u \to \infty} \frac 1 u \log \frac 1 u = \lim \limits _{u \to \infty} - \frac {\log u} u = 0 .$$

So, if $f_n (x) = \frac x n \log \frac x n$, we have obtained $f_n \to 0$.

To see whether the convergence is uniform, all we have to do is study whether

$$\sup \limits _{x \in (0,1)} |f_n (x) - 0| = 0 .$$

Since $f_n \le 0 \ \forall n \ge 1$, we must check whether $\sup \limits _{x \in (0,1)} g_n = 0$ where

$$g_n(x) = |f_n (x)| = - \frac x n \log \frac x n .$$

Let us explicitly find the $\sup$ of $g_n$ on $(0,1)$. We have that $$g_n ' = - \frac 1 n \log \frac x n - \frac x n \frac 1 {\frac x n} \frac 1 n = - \frac 1 n (\log \frac x n + 1) .$$

The critical point are the roots of $\log \frac x n + 1 = 0$, which means $\frac x n = \frac 1 {\rm e}$, which gives $x = \frac n {\rm e}$. Note that for $n \ge 3$ this point lives outside of $(0,1)$, so for $n \ge 3$ we have that $g_n ' \ge 0$, so $g_n$ is non-decreasing, therefore its $\sup$ must be reached in $1$ and be $- \frac 1 n \log \frac 1 n$.

Finally, note that $\lim \limits _{n \to \infty} - \frac 1 n \log \frac 1 n = 0$ (with the same technique as in the beginning of this answer), so $\sup \limits _{x \in (0,1)} g_n = 0$ as desired, so yes, the convergence is uniform.

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There is an alternative aproach to this problem, along the lines of your unachieved (but essentially correct) proof.

First, extend $f_n$ by continuity to a function $F_n$ defined on $[0,1]$: noting that $\lim \limits _{x \to 0, \ x > 0} f_n (x) = 0$ allows us to define $F_n (x) = \left\{ \begin{eqnarray} 0 &,& \quad x = 0 \\ \frac x n \log \frac x n &,& \quad x \in (0,1) \\ \frac 1 n \log \frac 1 n &,& \quad x = 1 \end{eqnarray} \right.$.

It is easy to see that $F_n$ is continuous, increasing ($F_n ' \ge 0$), and that $F_n \to 0$ pointwisely. This means that we may apply Dini's theorem and get that $F_n \to 0$ uniformly on $[0,1]$.

Finally, to show that $f_n \to 0$ uniformly on $(0,1)$ note that

$$0 \le \sup \limits _{x \in (0,1)} |f_n (x) - 0| \le \sup \limits _{x \in (0,1)} |F_n (x) - 0| \le \sup \limits _{x \in [0,1]} |F_n (x) - 0| \to 0$$

(where the last convergence encodes the fact that $F_n \to 0$ uniformly), so $f_n \to 0$ uniformly on $(0,1)$.

As you see, you were very close to the correct proof, your intuition was good but there were a few technical details missing.