Let $f \in L_1[-\pi,\pi]$ and let $\epsilon > 0$.
- Show that there exists a continuous $2\pi$ periodic function $g$ such that $\int_{[-\pi,\pi]}|f-g| < \epsilon$.
- Show that there exists a trigonometric polynomial $T$ such that $\int_{[-\pi,\pi]}|f-T| < \epsilon$.
I am just a bit confused about the specifics for proving part 1. I have:
Given $\epsilon > 0$ and $f \in L_1[-\pi,\pi]$, $\exists h \in C[-\pi, \pi]$ such that $\int_{[-\pi,\pi]}|f-h| < \epsilon/2$.
What I have for the latter half of part 1 is what's confusing me. I think my details are worked out incorrectly but I'm not sure.
I want to define $g(x) = h(x) \forall x \in [-\pi,\pi)$ and $g(\pi) = g(-\pi)$. Then extend this to a $2\pi$ periodic function.
However, this means $g$ is not necessarily continuous (could be a discontinuity as we approach $g(\pi)$). So I actually need to define it a bit differently.
My next idea was to take: $g(x) = h(x) \forall x \in [-\pi,\pi-\epsilon/2]$, $g(\pi) = g(-\pi)$, and linear from $g(\pi-\epsilon/2)$ to $g(\pi)$ then extend this to a $2\pi$ periodic function. But then:
$\int_{[-\pi,\pi]}|g-h| = \int_{[\pi-\epsilon/2, \pi]}|g-h| \leq k*\epsilon/2$
where $|g-h| \leq k$ and $m([\pi-\epsilon/2, \pi]) = \epsilon/2$.
But obviously this is also not what I want. Can I instead define $g(x) = h(x)$ over the interval $[-\pi,\pi-\epsilon/2k]$ where k is as above (and then proceed similarly)?
And part 2 is easy once we have part 1 (since continuous, $2\pi$-periodic functions can be uniformly approximated by trigonometric polynomials).