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Let's consider the sequence of functions $f_n:[0,1] \to \mathbb{R}$ , defined by $f_n(x) = \frac{nx}{1+n^2x^p}$ for $p>0$. I proved that this function converges pointwise to the constant function $f(x)=0$ . The question is about $p$ in which $p$, the convergence is uniform?

At least I know that if $1>p>0$ then $f_n$ is increasing ( it's because $f ' _n$ >0 in that case) But I don't know how to attack the other cases

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up vote 1 down vote accepted

Let's compute $f_n'$ (as you allready did) to compute $f_n$'s extreme values: \[ f_n'(x) = \frac{n(1 + n^2x^p) - n^3px^p}{(1 + n^2x^p)^2} \] So we have for $p \ne 1$: \begin{align*} f_n'(x) &= 0 \\ \iff n + n^3x^p - pn^3x^p &= 0\\ \iff (1-p)n^3x^p &= -n\\ \iff x^p &= \frac{1}{n^2(p-1)}\\ \end{align*} For $x \in [0,1]$, this is only possible if $p > 1$. So, for $p \le 1$, $f_n$ doesn't have internal extrema, hence \begin{align*} \|f_n\|_\infty &\le \max\{|f_n(0)|, |f_n(1)|\}\\ &= |f_n(1)|\\ &= \frac n{1+n^2} \to 0. \end{align*} For $p > 1$, a third value comes into play (at least for large $n$), namely \begin{align*} f_n\Bigl(\bigl(n^2(p-1)\bigr)^{-1/p}\Bigr) &= \frac{n\cdot \bigl(n^2(p-1)\bigr)^{-1/p}}{1 + n^2\bigl(n^2(p-1)\bigr)^{-1}}\\ &= \frac{n\cdot \bigl(n^2(p-1)\bigr)^{(p-1)/p}}{n^2(p-1) + n^2}\\ &= \frac{n^{1 + 2(p-1)/p}}{n^2}\cdot \frac{(p-1)^{(p-1)/p}}{p}\\ &= n^{1 - 1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p} \end{align*} Now for $n \in \mathbb N$ with $\frac 1{n^2} < p-1$ and $n^{1-1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p} \ge 1$ (which finally holds for all $n$ if $p > 1$), we have \[ \|f_n\|_\infty = n^{1 - 1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p} \to \infty. \]

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