Let $f$ be a Riemann integrable function on $[-\pi,\pi]$, and let $\epsilon>0$. Prove:
1) There is a function $g\in C[-\pi,\pi]$, satisfying $$\int_{-\pi}^{\pi}|f(x)-g(x)| \mathop{dx}<\epsilon$$ 2) There is a continuous, $2\pi$-periodic function $h\in C^{2\pi}$ satisfying}$$\int_{\pi}^{\pi}|f(x)-h(x)|\mathop{dx}<\epsilon$$
and 3) There is a trigonometric polynomial $T$ with $$\int_{-\pi}^{\pi}|f(x)-T(x)|\mathop{dx}<\epsilon$$
Some background and thoughts.
I am working on some textbook problems in the section on Trigonometric Polynomials. The following questions are very related so I hope it is okay to post all of them under one thread. The only theorem that has been discussed in this chapter is Weierstrass's Second Theorem which says that if I have a continuous function with period $2\pi$ and $\epsilon>0$, then there exists a trig. polynomial $T$, such that $\|f-T\|_\infty < \epsilon$.
I think that showing that any of the above are true is essentially showing that the difference of the two functions under the infinity norm is less than epsilon, which would result in the integral being less than epsilon as well. I'm not sure how to start 1) or 2), but my attempt at 3) is as follows:
We note that any linear combination of the basis functions of a trig polynomial has period $2\pi$ and is continuous. Then by Weierstrass' 2nd Theorem, there exists a $T$ such that $\|f-T\|_\infty<\epsilon$. But I don't think it would be that simple...
Any thoughts on how to solve these? Thanks for your help.
For 1), just use the assumption that $f$ is Riemann integratable. By integrablility, there exist a step function $s$ (just pick the Darboux lower sum), such that $$s(x)\leq f(x), \int_{-\pi}^{\pi}f(x)-s(x) \mathop{dx}<\frac{\epsilon}{2}$$
Then modify the jump discontinuity point of the step function (only finitely many of these points ) so that it becomes a continuous piecewise linear function $g$, such that $$\int_{-\pi}^{\pi}|s(x)-g(x)| \mathop{dx}<\frac{\epsilon}{2}$$ Then use the triangle inequity to compare $f$ and $g$.
For 2) from the function $g$ obtained in 1), modify the value of $g$ at the endpoint similarly such that $g(-\pi)=g(\pi)$. Then extend it to a continuous $2\pi$ periodic function $h$.
For 3) given integrable $f$, first by part 2), approximate it by continuous, $2\pi$ periodic $h$, then by Weierstrass theorem, find $T$ and note $$\int_{-\pi}^{\pi}|h(x)-T(x)|\mathop{dx}\leq \int_{-\pi}^{\pi}\|f(x)-T(x)\|_\infty\mathop{dx}$$