2
$\begingroup$

Let $f_n:[0,1]\to [0,1]$ continuous functions and let $f:[0,1]\to [0,1]$ such that $f_n$ converges uniformly to $f$. Show that $\frac{1}{n} \sum_{k=1}^n f_k$ also converges uniformly to $f$.

Now, I've seen a proof which starts with:

$$\frac{1}{n} \sum_{k=1}^n (f_k - f) = \frac{1}{n} \left[ f-f_1 + f-f_2 + \ldots + f_n -f \right] \le \ldots \le \varepsilon$$

BUT, why is it showing uniform converges? I mean, shouldn't it start with:

$$ \left( \frac{1}{n} \sum_{k=1}^n f_k \right) -f $$

$\endgroup$

marked as duplicate by AD., Dirk, Joel Reyes Noche, Martin Sleziak, Najib Idrissi Mar 7 '15 at 10:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The two things are same, as $f = \frac1n\sum_{k=1}^n f$. $\endgroup$ – Sten Mar 4 '15 at 12:55
2
$\begingroup$

Notice, that all sums have finite number of elements, therefore $$\left( \frac{1}{n} \sum_{k=1}^n f_k \right) -f = \left( \frac{1}{n} \sum_{k=1}^n f_k \right) - \frac{1}{n}\sum_{k=1}^{n}f = \frac{1}{n} \sum_{k=1}^n (f_k - f). $$

$\endgroup$
2
$\begingroup$

Hint: We have $\frac 1n \sum_{i=1}^n f = \frac 1n \cdot n f = f$, hence $$ \frac 1n \sum_{i=1}^n (f_i - f) = \frac 1n \sum_{i=1}^n f_i - \frac 1n \sum_{i=1}^n f = \frac 1n \sum_{i=1}^n f_i - f $$

$\endgroup$
2
$\begingroup$

The left hand side is equivalent to what you have. To show this, write $$(\frac 1 n \sum _{k=1} ^nf_k)-f=(\frac 1 n \sum _{k=1} ^nf_k)-(\frac 1 n)(n\cdot f)=(\frac 1 n \sum _{k=1} ^nf_k)-(\frac 1 n)\sum _{k=1} (nf)=\frac 1 n (\sum _{k=1} ^nf_k-\sum _{k=1} ^n f)= \frac 1 n \sum _{k=1} ^n (f_n -f)$$

Yeah distribution

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.