Does the sequence $\displaystyle g_n(t)=\int_0^1 \frac{t f(\sqrt[n]{x})}{1+t f(\sqrt[n]{x})} dx$ converge uniformly on $[0, \infty)$?

Question

Let $f:[0, 1] \to [0, \infty)$ be a continuous function such that $f(1)=0$. Does the sequence $\displaystyle g_n(t)=\int_0^1 \frac{t f(\sqrt[n]{x})}{1+t f(\sqrt[n]{x})} dx$ converge uniformly on $[0, \infty)$?

I dealt with the pointwise convergence by using DCT: if we denote $u_n(t, x)=\frac{t f(\sqrt[n]{x})}{1+t f(\sqrt[n]{x})}$, then for every $t\in [0, \infty)$ we will have $\displaystyle \lim_{n\to \infty}u_n(t, x)=0$ for $x\in (0, 1]$ and $|u_n(t, x)|\le 1$ for $x\in [0, 1]$ and $t\in [0, \infty)$, so we may use DCT to conclude that $\displaystyle \lim{n\to \infty}g_n(t)=0$ for every $t\in [0, \infty)$.

However, I want to know if this convergence is uniform. I tried looking at $\displaystyle \sup_{t\in [0, \infty)}g_n(t)$, but I couldn't make any progress.

Answer 1

No.

The convergence is not uniform.

Take the function $f(x)=1-x,$ then

$$\sup_{t \in [0,+\infty)}|g_n(t)|=1$$

Indeed if you fix and $n \in \Bbb{N}$ then $\lim_{k \to +\infty}g_n(k) \to 1 $ from DCT thus $$\sup_{t \geq 0}|g_n(t)| \geq \sup_{k \in \Bbb{N}}|g_n(t)|=1$$

Answer 2

I would use that the function \begin{equation} \frac{t}{1+t} \end{equation} is strictly increasing.

Then, note that for every $\epsilon>0$ there exists an $n>0$ such that for all $x\in[0,1]$ we obtain \begin{equation} 1-\epsilon\leq f(x^{1/n})\leq 1+\epsilon. \end{equation}

Now, since $\frac{t}{1+t}$ is strictly increasing note that \begin{equation} tf(x^{1/n})\leq t(1+\epsilon)\implies \frac{tf(x^{1/n})}{1+tf(x^{1/n})}\leq \frac{t(1+\epsilon)}{1+t(1+\epsilon)}. \end{equation}

The rest is straightforward.