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Let $f_n \in C_c^\infty(0,\infty)$ for $n\in \mathbb N$, $f: (0,\infty) \in L^p(0,\infty)$, where $1<p<\infty$ and $\|f_n-f\|_p \rightarrow 0 $ as $n\rightarrow \infty$.

We define $$ F_n(x)=\frac{1}{x} \int_0^x f_n(t)dt, $$ $$ F(x)=\frac{1}{x} \int_0^x f(t)dt. $$ Assuming that we know that $\|F_n\|_p \leq \frac{p}{p-1} \|f_n\|_p$ for $n\in \mathbb N$

(or more generally if necessarilly that this inequality holds for all functions from $C_c^\infty$), how to prove that $\|F_n-F\|_p \rightarrow 0$ as $n \rightarrow \infty$?

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  • $\begingroup$ I know that it is a consequence of the Hardy's inequality. But i my question we asume that we know this inequality only for continuous functions with compact suport. $\endgroup$ – Alex Oct 8 '14 at 17:03
  • $\begingroup$ I need only $p>1$. $\endgroup$ – Alex Oct 9 '14 at 6:14
  • $\begingroup$ Does not the claim follow trivially from the inequality $\|F_n\|_p\le \frac{p}{p-1}\|f_n\|_p$? Like $\|F_n-F\|_p \le \frac p{p-1}\|f_n-f\|$? After all, this integral operator is linear. $\endgroup$ – daw Oct 9 '14 at 6:37
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I'll consider only $p>1$.

Since $\{f_n\}$ is a Cauchy sequence in $L^p$ norm, apply Hardy's inequality to $f_n-f_m$ yields that $\{F_n\}$ is also Cauchy in the $L^p$ norm. Hence $F_n\to G$ in $L^p$ for some $G\in L^p$. It remains to show $F=G$.

Recall that $L^p$ convergence implies a.e. convergence of some subsequence. Thus, we have $F_{n_k}\to G$ pointwise. On the other hand, the fact that $f_n\to f$ in $L^p$ implies $\int_0^x f_n\to \int_0^x f$ for every $x$; this means $F_n\to F$ pointwise. Conclusion: $F=G$ a.e.

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  • $\begingroup$ I needed only $p>1$. Very thanks for answer! $\endgroup$ – Alex Oct 9 '14 at 6:16

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