19
$\begingroup$

The following is a proof in Brezis book. It shows the separability of $L^{p}$ spaces:

enter image description here

I have a few questions regarding the proof:

  1. It says 'it is easy to construct a function $f_{2} \in \varepsilon...$" and it also says " it suffices to split $R$ into small cubes...'. Would it work to choose $f_{2}$ in the following way:

    Assume we split $R$ as suggested. Let $R_{i}$ denote each small cube of $R$, consider $f_{2_{i}} := C_{i}\chi_{R_{i}}$ where $C_{i}$ is a constant chosen from $[0, \delta - (\text{sup} f|_{R_{i}} - \text{inf} f|_{R_{i}})$, then let $f_{2}(x) := \sum_{i}f_{2_{i}}(x)$. It would then follow that $\Vert f_{1} - f_{2} \Vert_{\infty} < \epsilon$. Is this fine?

  2. Can anyone see how the inequality $\Vert f_{1} -f_{2} \Vert_{p} \leq \Vert f_{1}-f_{2} \Vert_{\infty}|R|^{\frac{1}{p}}$ is obtained?

  3. Where exactly is the separability of $\Omega = \mathbb{R}^{N}$ used?

Note that $\chi$ denotes the characteristic function.

Thanks a lot for any assistance. Let me know if something is unclear.

$\endgroup$
  • 5
    $\begingroup$ The second one is rather easy: $\| f_1 - f_2 \|_p = \left ( \int_R |f_1 - f_2|^p \right )^{1/p} \leq \left ( \int_R \| f_1 - f_2 \|_\infty^p \right )^{1/p} = \left ( |R| \| f_1 - f_2 \|_\infty^p \right )^{1/p} = |R|^{1/p} \| f_1 - f_2 \|_\infty$. $\endgroup$ – Ian Jul 26 '14 at 19:10
  • $\begingroup$ Okay and as I was also shown 2, should be: $C_i \in [\inf f_1\lvert_{R_i}, \sup f_1\lvert_{R_i}]$. I have $\lVert f_2\rVert_\infty < \delta$, not $\lVert f_1-f_2\rVert_\infty < \delta$. $\endgroup$ – Lucio D Jul 26 '14 at 19:59
13
$\begingroup$
  1. As you mentioned in a comment, you should choose $C_i \in \Big[ \inf f_1 \restriction R_i, \sup f_1 \restriction R_i \Big]$. This way you get $\big\|f_1 -f_2\big\|_\infty < \delta$. An adequate choice of $\delta > 0$ (that is $\delta |R|^{\frac{1}{p}} < \varepsilon$) would then give you $\big\| f_1 - f_2\big\|_p \leq \varepsilon$.
  2. $$ \big\| f_1 - f_2\big\|_p = \left(\int_R |f_1 - f_2|^p \right)^{\frac{1}{p}} \\ \leq \left( \int_R \big\| f_1 - f_2\big\|_\infty^p \right)^{\frac{1}{p}} \\ \left(|R|\cdot \big\| f_1 - f_2\big\|_p \right)^{\frac{1}{p}} \\ |R|^{\frac{1}{p}} \cdot \big\| f_1 - f_2\big\|_\infty.$$
  3. A metrizable space is separable if and only if it is second-countable. This means that the euclidean topology on $\mathbb{R}^n$ has a countable basis. This countable basis is explicitely described in the first paragraph of the proof and is put to use in the second paragraph.
$\endgroup$
0
$\begingroup$

As you tell: As you mentioned in a comment, you should choose $C_i \in \left[ \inf (f_1|_{R_i}), \sup(f_1|_{R_i})\right]$. This way you get $\| f_1 − f_2 \|_{\infty} < \delta$. An adequate choice of $\delta > 0$ (that is $\delta | R |^{\frac{1}{p} } < \varepsilon$) would then give you $\| f_1 − f_2 \|_{p} \le \varepsilon$. Can Anyone tell me how I can obtain this inequality $\| f_1 − f_2 \|_{\infty} < \delta$. I understand the proof but I dont know how to get this inequality. Best regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.