A difficulty in understanding the new function defined in the approximation lemma(1.5) in Stein & Shakarachi Fourier Analysis.

Question

I did not understand the paragraph that says "Now we can modify $f^{*}$ to make it continuous and periodic yet still approximate $f$ in the sense of the lemma....... below to clarify the situation", could anyone explain this for me?

Answer 1

The intuition to what you do is you start with $f^*$ (the approximation by step functions), and at the very end of one of the "steps" you connect that step to the next one by a straight line. This will make it continuous (as it's piecewise linear), and if you do it "close enough to the end" it'll still be very near to $f$. This technique is very common in analysis.

Then, we have $\tilde{f}$ is defined on $[-\pi,\pi]$, and you can extend it to all of $\mathbb R$ by just ensuring the extension is $2\pi$-periodic.

Regarding getting the bound before equation $2$:

The idea is that we know that $f^*$ consists of $N-1$ step functions. For each step function, we have that outside of a $\delta$ ball around the discontinuities that $\tilde{f} = f^*$, so their difference is $0$. Inside that $\delta$-ball at the discontinuity, we (for the sake of getting any bound) say that $|\tilde{f}-f^*| \leq 2B$ (as $B$ is a bound for both $f^*$ and $\tilde{f}$. This is just triangle inequality).

So, outside of each $\delta$ ball, we have that $|\tilde{f}-f^*| = 0$, and inside each $\delta$-ball we have that $|\tilde{f}-f^*| \leq 2B$. Let $B_\delta$ be the set of all of our $\delta$-balls. So, $B_\delta = \{B_\delta(p_1), B_\delta(p_2),\dots\}$,is a set of $\delta$-balls at discontinuity points. Let $[-\pi,\pi]\setminus B_\delta$ be "the rest of the interval". We have that: \begin{align*} \int_{-\pi}^\pi |\tilde{f}(x)-f^*(x)|dx & \leq \sum_{b\in B_\delta} \int_{b} |\tilde{f}(x)-f^*(x)|dx +\int_{[-\pi,\pi]\setminus B_\delta}|\tilde{f}(x)-f^*(x)|dx \\ & \leq\sum_{b\in B_\delta}\int_b|\tilde{f}(x)-f^*(x)|dx +\int_{[-\pi,\pi]}0dx \end{align*} Here, we used that $\tilde{f}(x) = f^*(x)$ outside of the $\delta$-balls. So, we just have to bound $\sum_{b\in B_\delta}\int_b|\tilde{f}(x)-f^*(x)|dx$. Because of our earlier bound $|f^*(x)-\tilde{f}(x)|\leq 2B$ inside the $\delta$-balls, we have that: $$\sum_{b\in B_\delta}\int_b |f^*(x)-\tilde{f}(x)|dx \leq \sum_{b\in B_\delta}\int_b 2Bdx = \sum_{b\in B_\delta} 2B(\text{diam}(b)) = \sum_{b\in B}2B(2\delta) = (N-1)2B(2\delta)$$ The bound we wanted was $2B(2\delta)N$. This is also a bound of what I found, but it's likely I made some small error and the direct result you should get is $2B(2\delta)N$.

So, it is just triangle inequality, as you say.