Let $F_n$ be a real valued function, $$F_n(r) = \frac{r}{n}(\psi(n) - \psi(r)$$ where $\psi$ is a digamma function. Let the sequence of functions $\{g_n\}$ be defined by $g_n(x) := F_n(nx)$.
Show that $\{g_n\}$ converges uniformly on [0,1] and to the function $g(x) := -xlog(x)$.
I am trying to understand the paper, "A variant of the Secretary Problem: the Best or the Worst" by L. Bayon, J. Grau, A. M. Oller-Marcen, M. Ruiz, P.M. Suarez.
It says there that this can be seen with little effort but I am at a loss.
Any hints on how I would prove this would be much appreciated.
We have
$$\tag{1}\begin{align}|F_n(nx) -(-x \log x)| &= |x(\psi(n) - \psi(nx)) + x\log x| \\&= |x(\psi(n) - \log n) - x(\psi(nx) - \log(nx)| \\ &\leqslant|x||\psi(n) - \log n| + |x (\psi(nx) - \log(nx))|\end{align}$$
There are many well-known representations of the digamma function. Using a recurrence relation we get following representation for integer arguments:
$$\psi(n) = \sum_{k=1}^{n-1} \frac{1}{k} - \gamma,$$
where $\gamma$ is the Euler-Mascheroni constant. With $0 \leqslant x \leqslant 1$, we then have $$\tag{2}|x||\psi(n) - \log n| \leqslant |\psi(n) - \log n| = \left| \sum_{k=1}^{n-1}\frac{1}{k} - \log n - \gamma\right| $$
From the definition of $\gamma$, the RHS of (2) converges to $0$ as $n \to \infty$ and , thus, for any $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$, independent of $x$, such that for all $n \geqslant N_1$, we have for the first term on the RHS of (1),
$$|x||\psi(n) - \log n| < \epsilon/2$$
Using Gauss' integral representation we have
$$\psi(z) = \log z - \frac{1}{2z} - \int_0^\infty \left(\frac{1}{2} - \frac{1}{t} + \frac{1}{e^{t} + 1} \right)e^{-tz} \, dt,$$
and it can be shown (see here) that for all $z > 0$,
$$\frac{1}{2z} \leqslant \log z - \psi(z) \leqslant \frac{1}{z}$$
Thus, for the second term on the RHS of (1) we have
$$\frac{1}{2n} = \frac{x}{2nx} \leqslant |x (\psi(nx) - \log(nx))| = x(\log (nx) - \psi(nx))\leqslant \frac{x}{nx} = \frac{1}{n},$$
and so for any $\epsilon > 0$ there exists $N_2 \in \mathbb{N}$, independent of x, such that for all $n \geqslant N_2$ and all $x \in (0,1]$ we have
$$|x (\psi(nx) - \log(nx))| < \epsilon /2$$
Thus, for all $n \geqslant (N_1,N_2)$ and all $x \in (0,1]$, we have
$$|F_n(nx) - (-x\log x)| < \epsilon$$
This establishes uniform convergence on $(0,1]$. To completely finish here, we would need to establish that $F_n(0) = 0$ by extension as is the case for $x\log x$ where we have $x \log x \to 0$ as $x \to 0$.