Prove that $a_n=\sqrt[n]{f\left(\frac{1}{n} \right)^n+f\left(\frac{2}{n} \right)^n+\dots+f\left(\frac{n}{n} \right)^n}$ is convergent

Question

$f$ takes positive values and is uniformly continuous. Prove that $$a_n=\sqrt[n]{f\left(\frac{1}{n} \right)^n+f\left(\frac{2}{n} \right)^n+\dots+f\left(\frac{n}{n} \right)^n}$$ is convergent

By uniform continuity, for large enough $n$ we have $|f\left(\frac{k}{n}\right)-f\left(\frac{k+1}{n} \right)|<\epsilon$. I tried to use this in order to bound the sum, but that power $n$ makes it too large...

Answer 1

If $f[0,1]\to [0,\infty)$, continuous, then $$ \lim_{n\to\infty}\sqrt[n] {\, f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+ f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+ f\left(\frac{n}{n}\right)^{n^{\phantom{|}}} }=\max_{x\in [0,1]}f(x). $$

Proof. Let $x_0\in[0,1]$, such that $$f(x_0)=\max_{x\in [0,1]}f(x)=M$$ and $\varepsilon>0$ arbitrary. Then there is a $\delta>0$, such that if $|x-x_0|<\delta$, then $\,|\,f(x)-f(x_0)|<\varepsilon$. Thus, if $$ n\ge n_1=\left\lfloor\frac{1}{\delta}\right\rfloor+1\quad\Longrightarrow\quad \frac{1}{n}<\delta, $$ and hence, for $n\ge n_1$, the distance of one of the points $\frac{1}{n},\frac{2}{n},\ldots,\frac{n}{n}$ is less that $\delta$ and thus among the values $f\left(\frac{1}{n}\right),f\left(\frac{2}{n}\right),\ldots,f\left(\frac{n}{n}\right)$, there exists one, say $f\left(\frac{k}{n}\right)$, which differs less that $\varepsilon$ from $f(x_0)$, i.e. $$ M-\varepsilon< f\left(\frac{k}{n}\right)\le M. $$ Next, as $Mn^{1/n}\to M$, there exists an $n_2$, such that if $n\ge n_2$, then $Mn^{1/n}<M+\varepsilon$.

Thus, for every $n\ge n_0=\max\{n_1,n_2\}$ $$ (M-\varepsilon)^n<f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+ f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+ f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}\le nM^n $$ and thus $$ M-\varepsilon<\sqrt[n]{f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+ f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+ f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}}\le Mn^{1/n}<M+\varepsilon $$ Altogether, for every $\varepsilon>0$, there exists an $n_0\in\mathbb N$, such that $$ n\ge n_0\quad\Longrightarrow\quad \left|\sqrt[n]{f\left(\frac{1}{n}\right)^{n^{\phantom{-}}}+ f\left(\frac{2}{n}\right)^{n^{\phantom{|}}}+\cdots+ f\left(\frac{n}{n}\right)^{n^{\phantom{|}}}}-M\right|<\varepsilon. $$ QED

Answer 2

Hint: Well, if we have a finite number of positive terms $a_1,a_2,\ldots,a_m$ then $$ \lim_{n\to +\infty}\sqrt[n]{a_1^n+a_2^n+\ldots+a_m^n} = \max(a_1,a_2,\ldots,a_m)$$ is well-known and not difficult to prove. If $f$ is positive and continuous on $[0,1]$, by letting $M=\sup_{[0,1]}f(x)=\max_{[0,1]}f(x)$ we have $$(M-\varepsilon_n)^n\leq f\left(\tfrac{1}{n}\right)^n+f\left(\tfrac{2}{n}\right)^n+\ldots+ f\left(\tfrac{n}{n}\right)^n \leq n M^n$$ and $\lim_{n\to +\infty}\sqrt[n]{n}=1$, so the situation is essentially the same.

Answer 3

Let $M=\max f|_{[0,1]}$. Then for every $\epsilon>0$, we find an interval of positive length $r$ such that $f(x)>M-\frac\epsilon2$ in that interval. Note that at least $nr-1$ of the points fall into that interval. We conclude $$ nM^n\ge f(\tfrac1n)^n+\ldots +f(\tfrac nn)^n\ge (nr-1)(M-\tfrac\epsilon2)^n$$ and so $$ \sqrt[n] n\cdot M\ge a_n\ge\sqrt[n]{nr-1}\cdot (M-\tfrac \epsilon2).$$ As $\sqrt[n] n\to 1$ and $\sqrt[n]{nr-1}\to 1$, we conclude that, say, $|a_n-M|<\epsilon$ for almost all $n$, i.e., $a_n\to M$.