Proof of separability of $L^p$ spaces

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

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.

• The second one is rather easy: $\| f_1 - f_2 \|_p = \left ( \int_R |f_1 - f_2|^p \right )^{1/p} \leq \left ( \int_R \| f_1 - f_2 \|_\infty^p \right )^{1/p} = \left ( |R| \| f_1 - f_2 \|_\infty^p \right )^{1/p} = |R|^{1/p} \| f_1 - f_2 \|_\infty$.
– Ian
Commented Jul 26, 2014 at 19:10
• 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 $\lVert f_1-f_2\rVert_\infty < \delta$. Commented Jul 26, 2014 at 19:59
• Why isn't the answer accepted? Commented Mar 30, 2021 at 12:44
• Why $\|f_1-f_2\|_\infty<\epsilon$?
– Math
Commented Feb 15, 2022 at 13:50

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

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

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