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Let $F: [0,1]^n \to \mathbb{R}$ be a Riemann integrable function. Let $$ J(H) =\int_0^1.. \int_0^1 (\int_{[0,1]} H(t_1, t_2, ..., t_n) dt_1)^2 dt_2 ... dt_n. $$ Given $\epsilon > 0$ I want to find $G$ a smooth function such that $$ |J(F) - J(G)| < \varepsilon. $$ I was wondering how I could start with this. Any comments are appreciated! Thank you!

Edit: My apologies for the confusion! I had mistakenly oversimplified the part of the article I am trying to understand.

Let $$ J^{(m)}(H) =\int_0^1.. \int_0^1 (\int_{[0,1]} H(t_1, t_2, ..., t_n) dt_m)^2 dt_1 ... dt_{m-1} dt_{m+1}...dt_n. $$ $$ I(H) = \int_0^1.. \int_0^1 H^2(t_1, t_2, ..., t_n) dt_1 ...dt_n. $$

Let $$ M = \sup_{F \in S} \frac{\sum_{m=1}^nJ^{(m)} (F)} {I(F)}, $$ where $S$ is the set of all Riemann integrable functions on $[0,1]^n$. Let $\delta > 0$ small. Then there exists $F_0 \in S$ such that $$ \sum_{m=1}^nJ^{(m)} (F_0) > (M- \delta) I(F_0) > 0. $$ This I understand. But then they claim that since $F_0$ is Riemann integrable, there exists $F_1$ a smooth function such that $$ \sum_{m=1}^nJ^{(m)} (F_1) > (M- 2\delta) I(F_1) > 0. $$ I was wondering how this follows. I had mistakenly thought that if I could understand what I had asked above then I could deduce this, but I don't think this is the case seeing the comments and answers. Any explanation for this would be appreciated! Thank you!

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    $\begingroup$ I believe you mean $|J(F-G)| < \varepsilon$, (otherwise you can just take $G$ to be a constant function of value $J(F)$, which is only a very rough approximation of $F$). $\endgroup$
    – Joel Cohen
    Aug 4 '19 at 15:49
  • $\begingroup$ Thank you for the comments/answers! I have fixed the question. $\endgroup$ Aug 5 '19 at 11:52
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As written, is utterly trivial. You can take a constant (what constant?) $G$.

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  • $\begingroup$ Let me fix the question... $\endgroup$ Aug 5 '19 at 11:41
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If the limits can vary, you can't. Take e.g. the function that is 1 up to $1/2$, 2 afterwards. No smooth function can have such a derivative.

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