# Cauchy sequences when $p$ is a function

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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Where did you see this result? – Davide Giraudo Jul 23 '12 at 21:51
From some qualifying exams in Analysis in the USA. – checkmath Jul 23 '12 at 23:34
With $f$ being a step function - did you also use the completeness of $L^p$? – vanguard2k Jul 24 '12 at 6:23
Maybe the fact that $p$ can be approximate uniformly by step functions can help. – Davide Giraudo Jul 24 '12 at 10:37
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 – checkmath Jul 24 '12 at 17:07

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 $\lim_{n\to +\infty}\int_{\Bbb R^d}|f(x)-f_{\varphi(n)}(x)|^{p(x)}dx=0$.
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.
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. – Davide Giraudo Jul 26 '12 at 8:19