I am going through some proofs on dense subspaces of $L^p$ and I can't see where a certain inequalities came from and also why a certain step was taken. My question relates to the last expression on page 2 of these notes, along with the first line of page 3. I can see that the integral was split into two parts.
- I don't see where the last term on the right side of this expression comes from: $$\Vert f \ast K_\delta - f \Vert_p \le \varepsilon \int_{|y|<\eta}|K_\delta(y)|dy + 2\int_{|y|>\eta}\Vert f \Vert_p|K_\delta(y)|dy.$$ Specifically, where did the $2\Vert f \Vert_p$ come from?
- On the next page, it says the 'theorem follows by letting $\delta \to 0$, and then $\varepsilon \to 0$. I don't see how letting $\delta \to 0$ sends either of the two terms on the right hand side of (1.) to $0$. Furthermore, if it does send the second term to zero, why doesn't it also send the first term to zero...why do we specifically have to let $\varepsilon$ also go to zero?
Ad 1.
It says in the notes that $\| f * K_\delta - f \|_p \leq \int \| \tau_{-y} f - f \|_p |K_\delta(y)| dy$.
Estimate $\| \tau_{-y} f - f \|_p \leq \| \tau_{-y} f \|_p + \| f \|_p = 2 \|f\|_p$.
Ad 2.
$K_\delta$ has its mass concentrated on the ball of size $O(\delta)$. Once $\delta << \eta$, the mass outside the $\eta$-ball is very small. See condition (5) on the first page of the notes. The mass inside the $\eta$-ball is still there: $\lim_{\delta \searrow 0} \int_{|y| < \eta} |K_\delta(y)| dy = 1$.
So, taking the limit $\delta \searrow 0$ you are left with $\lim_{\delta \searrow 0} \| f * K_\delta - f \|_p \leq \epsilon$.