# Show that $\frac{1}{n} \sum_{k=1}^n f_k$ converges uniformly to $f$. [duplicate]

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:

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

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

• The two things are same, as $f = \frac1n\sum_{k=1}^n f$. – Sten Mar 4 '15 at 12:55
Notice, that all sums have finite number of elements, therefore $$\left( \frac{1}{n} \sum_{k=1}^n f_k \right) -f = \left( \frac{1}{n} \sum_{k=1}^n f_k \right) - \frac{1}{n}\sum_{k=1}^{n}f = \frac{1}{n} \sum_{k=1}^n (f_k - f).$$
Hint: We have $\frac 1n \sum_{i=1}^n f = \frac 1n \cdot n f = f$, hence $$\frac 1n \sum_{i=1}^n (f_i - f) = \frac 1n \sum_{i=1}^n f_i - \frac 1n \sum_{i=1}^n f = \frac 1n \sum_{i=1}^n f_i - f$$
The left hand side is equivalent to what you have. To show this, write $$(\frac 1 n \sum _{k=1} ^nf_k)-f=(\frac 1 n \sum _{k=1} ^nf_k)-(\frac 1 n)(n\cdot f)=(\frac 1 n \sum _{k=1} ^nf_k)-(\frac 1 n)\sum _{k=1} (nf)=\frac 1 n (\sum _{k=1} ^nf_k-\sum _{k=1} ^n f)= \frac 1 n \sum _{k=1} ^n (f_n -f)$$