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

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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.

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    $\begingroup$ 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
    Commented Jul 26, 2014 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
    Commented Jul 26, 2014 at 19:59
  • $\begingroup$ Why isn't the answer accepted? $\endgroup$ Commented Mar 30, 2021 at 12:44
  • $\begingroup$ Why $\|f_1-f_2\|_\infty<\epsilon$? $\endgroup$
    – Math
    Commented Feb 15, 2022 at 13:50

2 Answers 2

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  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.
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  • $\begingroup$ Why with this choice of $C_i$ we have $\|f_1-f_2\|_\infty<\delta$? $\endgroup$
    – Math
    Commented Feb 15, 2022 at 12:52
  • $\begingroup$ Consider $f_{2_i} := C_i \chi_{R_i}$ and $f_2 = \sum\limits_{i} f_{2_i}(x)$. Then, $\|f_1 - f_2\|_\infty = \sup\limits_{x \in R}\big|f_1(x) - f_2(x)\big| = \max\limits_{i}\sup\limits_{x \in R_i} \big|f_1(x) - f_{2_i}(x)\big| = \max\limits_{i}\sup\limits_{x \in R_i} \big|f_1(x) - C_i\big| \leq \max\limits_{i} \big| \sup f_i \restriction R_i - \inf f_i \restriction R_i\big| \leq \max\limits_{i} \delta = \delta$. $\endgroup$
    – M.G
    Commented Mar 24 at 11:25
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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

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  • $\begingroup$ See my comment above. $\endgroup$
    – M.G
    Commented Mar 24 at 11:27

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