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Consider $p$ being a positive bounded and measurable function and $\{f_k\}$ a sequence satisfying

$$\int_{R^d} |f_k(x)|^{p(x)}dx<\infty$$ and $$\lim_{m,j\to \infty}\int_{R^d} |f_j(x)-f_m(x)|^{p(x)}dx=0$$

then there is a $f$ such that

$$\int_{R^d} |f(x)|^{p(x)}dx<\infty$$ and $$\lim_{m\to \infty}\int_{R^d} |f(x)-f_m(x)|^{p(x)}dx=0.$$

I proved the result for $p$ being a step function but I couldn't extend the result to the general case.

Other way was to use a mimesis of that $L^p$ is complete.

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  • $\begingroup$ Where did you see this result? $\endgroup$ – Davide Giraudo Jul 23 '12 at 21:51
  • $\begingroup$ From some qualifying exams in Analysis in the USA. $\endgroup$ – checkmath Jul 23 '12 at 23:34
  • $\begingroup$ With $f$ being a step function - did you also use the completeness of $L^p$? $\endgroup$ – vanguard2k Jul 24 '12 at 6:23
  • $\begingroup$ Maybe the fact that $p$ can be approximate uniformly by step functions can help. $\endgroup$ – Davide Giraudo Jul 24 '12 at 10:37
  • $\begingroup$ If $p$ is bounded by 1 the result is ipsis litteris the result for L^p with p constant. but for p greater than is not so trivial $\endgroup$ – checkmath Jul 24 '12 at 17:07
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Let $\varphi\colon \Bbb N\to \Bbb N$ a (strictly) increasing map such that for each integer $n$ $$\int_{\Bbb R^d}|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}dx\leq \frac 1{4^n}.$$ Let $A_n:=\{x:|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}>1/2^n\}$. We have $\mu(A_n)\leq \frac 1{2^n}$, hence for almost every $x$, we can find $N(x)$ such that for all $n\geq N(x)$, $|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}\leq \frac 1{2^n}$. Since $p>0$, the sequence $\{f_{\varphi(n)}\}$ is almost everywhere convergent, say to a function $f$. We know that for $\alpha\in (0,1)$, we have, since $t\mapsto t^{\alpha}$ is sub-additive,
$$|a+b|^{\alpha}\leq |a|^{\alpha}+|b|^{\alpha},$$ and denoting $M:=\sup_{t\in\Bbb R^d}p(t)$, we have when $p\geq 1$, by convexity, $$|a+b|^{p(x)}\leq 2^{p(x)-1}(|a|^{p(x)}+|b|^{p(x)})\leq 2^{M-1}(|a|^{p(x)}+|b|^{p(x)}).$$ With these inequalities and Fatou's lemma, we can see that $\int_{\Bbb R^d}|f(x)|^{p(x)}$ is finite and $\lim_{n\to +\infty}\int_{\Bbb R^d}|f(x)-f_{\varphi(n)}(x)|^{p(x)}dx=0$.

The two mentioned inequalities show that the whole sequence converges.

It seems we didn't use any special feature of $\Bbb R^d$, and the result is true if we replace it by any set, and the Lebesgue measure by any positive one.

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  • $\begingroup$ As I told it was only an intelligent use of the proof that L^p is complete. $\endgroup$ – checkmath Jul 25 '12 at 23:22
  • $\begingroup$ Actually yes, and the two inequalities of convexity. It's an interesting exercise. Maybe there is an easy counter-example when $p$ is not bounded. $\endgroup$ – Davide Giraudo Jul 26 '12 at 8:19

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