Theorem: Let $ f \in R([0,2 \pi]) $. Then, $ \forall \epsilon>0 $ there exists a continuous function $ g $ on $ [ 0, 2 \pi ] $ s.t. $ || f - g || < \epsilon $.
Attempt ( inspired from lecture notes and posts in forum [ links below ] ):
Denote $ M = \sup_{[0,2 \pi]} f $ and let $ \epsilon >0 $. Then ( Usage of Darboux's definition for integrability ) there exists $ \delta > 0 $ . We'll choose a partition $ \prod = ( x_0 < x_1 < \cdots < x_n ) $ of $ [ 0,2 \pi ] $ s.t. $ \lambda(\prod) < \delta $ and we'll have that $ \omega(f,\prod) \leq \epsilon /M $. For every $ i \leq n $, we'll define $ g $ on the interval $ [ x_{i-1},x_i] $ from our partition as such: $ \forall x \in [x_{i-1},x_i] $ ,
$
g(x) = \begin{cases}
f(x_1) & x=x_1 \\
f(x_2) & x=x2 \\
f(x) & \text{otherwise}\end{cases}
$
Note that $ g $ is continuous. Also,
$ \forall x \in [x_{i-1},x_i] $ , $ | f(x) - g(x) | \leq \sup_{[x_{i-1},x_i]} f - \inf_{[x_{i-1},x_i]} f = \omega(f,[x_{i-1},x_i]) $ [here I stopped ]
From the point where I stopped the lecture notes say that
$ || f - g || = \frac{1}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) |^2 dx \leq \frac{2M}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) | dx \leq \frac{M}{\pi} \sum_{i=1}^{K} \omega(f,[x_{i-1},x_i]) \triangle{x_i} = \frac{M}{\pi} \omega(f,\prod) \leq \epsilon/3 < \epsilon $
and the proof's finished.
Question:
- I don't understand the transition " $ \frac{1}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) |^2 dx \leq \frac{2M}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) | dx $ " , how did we get to $ \frac{2M}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) | dx $?
- I don't understand the transition " $ \frac{2M}{2\pi} \int_{0}^{2\pi} | f(x) - g(x) | dx \leq \frac{M}{\pi} \sum_{i=1}^{K} \omega(f,[x_{i-1},x_i]) \triangle{x_i} $ " , how did we get to $ \frac{M}{\pi} \sum_{i=1}^{K} \omega(f,[x_{i-1},x_i]) \triangle{x_i} $?
- Is it required that $ f(0) = f( 2 \pi) $? is this used somewhere in the proof above?
Notes about notation:
- $ f \in R([a,b]) $ means $ f$ is Riemann integrable on $[a,b] $ .
- $ \lambda(\prod) = max_{i=1,...,n}|{ \triangle x_i}| $ ( this is the mesh of partition $ \prod $ )
- $\omega(f, J)=\sup _{J} f-\inf _{J} f=\sup _{x, y \in J}(f(x)-f(y))$, where J is an interval.
- $\omega(f, \Pi)=\sum_{i=1}^{n} \omega\left(f,\left[x_{i-1}, x_{i}\right]\right) \Delta x_{i}$ , where $ \Pi $ is a partition of some closed interval.
The links Show that there is a continuous function $g$ on $[-\pi,\pi]$ satisfying $||f-g||_{2} < \epsilon$ and Show there is some continuous, $2\pi$ periodic function $h$ such that $||f-h||_2<\epsilon$ helped me until I reached the inequality transitions.