Given a function $f:[0,2]\to \mathbb{R}$ that is continuous and bounded on $[0,2]$, show that the function $f_n(x):[0,1]\to \mathbb{R}$ where $$ f_n(x) = \frac 1n \sum_{k=0}^{n-1} f\left(x + \frac kn\right)$$ is uniformly convergent.

So far I believe that $\left(f_n(x)\right) \to\int_x^{x+1} f(x)\;\text{dx}$ and I've broken it down as $$ \begin{align} |f_n(x) - f(x)| & = \left|\frac 1n \sum_{k=0}^{n-1} f\left(x + \frac kn\right) - \int_x^{x+1}f(x) \; \text{dx}\right| \\ & = \left|\frac 1n \left(f(x) + f(x + 1/n) + ... + f\left(x+\frac{n-1}{n}\right)\right) - \int_{x}^{x+1/n} f(x) \; \text{dx} - \int_{x+1/n}^{x+2/n} f(x) \; \text{dx} - ... - \int_{x+(n-1)/n}^{x+1} f(x) \; \text{dx} \right|\\ & \leq \left|\frac{1}{n} f(x) - \int_{x}^{x+1/n} f(x) \; \text{dx} \right|+\left|\frac{1}{n} f(x+1/n) - \int_{x+1/n}^{x+2/n} f(x) \; \text{dx} \right|+...+\left|\frac{1}{n} f(x+\frac{n-1}{n}) - \int_{x+(n-1)/n}^{x+1} f(x) \; \text{dx} \right| \end{align} $$ Now I believe its a common result that $\left(\int_{a}^{a+1/n}f(x) dx\right) \to f(a)$ however that means I'm left with $$ |f_n(x) - f(x)| \leq \left|\frac{1}{n} f(x) - f(x) \right|+\left|\frac{1}{n} f(x+1/n) - f(x) \right|+...+\left|\frac{1}{n} f(x+\frac{n-2}{n}) - f(x) \right|+\left|\frac{1}{n} f(x+\frac{n-1}{n}) - f(x) \right|. $$ I'm not sure where to go from here. I'm also unsure whether or not if some of the integrals go to $f(x)$ or $f(x+1)$. I was also thinking proving absolute convergence since it seems like the sum can be broken into finite monotonic parts?


It is not true that $\int_a^{a+\frac 1n} f(t)dt \to f(a)$. Write $\frac 1 nf(x+\frac k n) -\int\limits_{x+\frac {k-1} n}^{x+\frac k n} f(t)dt$ as $\int\limits_{\frac {k-1} n}^{\frac k n} [f(x+\frac k n)-f(x+s)]ds$ (where I have used the substitution $s=t-x$) and use uniform continuity of $f$.

  • $\begingroup$ Ah yeah that's what I expected. in future for clarity if you perform a change of variables could you make it explicit the as there's a bit of ambiguity to your solution. $\endgroup$ Nov 21 '20 at 12:24
  • $\begingroup$ @JohnMiller I have edited. $\endgroup$ Nov 21 '20 at 12:29
  • $\begingroup$ Also what do you mean uniform continuity of $f$, I only said $f$ was bounded and continuous. $\endgroup$ Nov 21 '20 at 12:46
  • $\begingroup$ Any continuous function on $[0,2]$ (or $[a,b]$) is automatically bounded and uniformly continuous. Uniformly continuous means for any $\epsilon >0$ there exists $\delta >0$ such that $|f(x)-f(y)| <\epsilon$ whenever $|x-y| <\delta$. @JohnMiller $\endgroup$ Nov 21 '20 at 12:51
  • $\begingroup$ Ah yes sorry forgot we had a closed domain. $\endgroup$ Nov 21 '20 at 13:07

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