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If I have a sequence $\{f_n\}_{n\in \mathbb{N}} \subset L_p\cap L_q$ that is Cauchy under both norms, I am wondering if $f_n \to f$ in $L_p$ implies that $f_n\to f$ in $L_q$.

I have been working on this but I can't seem to figure out a definite proof.

Thank you!

PS: $1\leq p<q <\infty$

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Yes, if the sequence is a Cauchy sequence in both norms, it converges to the same function in both norms. That follows from the fact that a sequence converging in $L^r,\; 1 \leqslant r \leqslant \infty$, has a subsequence that converges almost everywhere to the limit function.

Since the $L^r$ spaces are complete, the condition that $(f_n)$ be a Cauchy sequence in both norms asserts the existence of limits in both spaces,

$$f_n \xrightarrow[n\to\infty]{L^p} f^{(p)};\qquad f_n \xrightarrow[n\to\infty]{L^q} f^{(q)}.$$

We can now extract a subsequence $(f_{n_k})$ that converges pointwise almost everywhere to the $L^p$-limit $f^{(p)}$. That subsequence is still a Cauchy sequence in both norms, and converges to $f^{(p)}$ resp. $f^{(q)}$ in the two spaces.

Since $f_{n_k} \xrightarrow[k\to\infty]{L^q} f^{(q)}$, we can extract a subsequence $(f_{n_{k_m}})$ of $(f_{n_k})$ that converges pointwise almost everywhere to $f^{(q)}$.

Then the subsequence $(f_{n_{k_m}})$ of $(f_n)$ converges pointwise almost everywhere to $f^{(p)}$, since it is a subsequence of $(f_{n_k})$, and it converges pointwise almost everywhere to $f^{(q)}$. Thus we have $f^{(p)} = f^{(q)}$ almost everywhere, and the $L^p$-limit is the same as the $L^q$-limit (as a class of measurable functions modulo the equivalence relation of being equal almost everywhere).

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  • $\begingroup$ But the first subsequence converges wrt the $Lp$ norm while the second converges wrt the $Lq$ norm. How does that help? $\endgroup$ – Merry Feb 26 '14 at 10:39
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    $\begingroup$ We have the premise that the entire sequence $(f_n)$ is a Cauchy sequence with respect to both norms. Then every subsequence is also a Cauchy sequence with respect to both norms. The $L^r$ spaces are all complete, so the sequence converges to $f^{(p)}$ in $L^p$, and to $f^{(q)}$ in $L^q$, and every subsequence also has that property. Now we extract a subsequence that also converges pointwise almost everywhere to $f^{(p)}$. That subsequence still converges in both norms to the respective limit. Now extract a subsequence of the first subsequence converging pointwise a.e. to $f^{(q)}$. $\endgroup$ – Daniel Fischer Feb 26 '14 at 10:54
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    $\begingroup$ Thank you! I wasn't aware of the fact that for any sequence convergent in the $L_p$ norm, there is a subsequence that converges pointwise a.e. That's a huge fact I missed. So I went and looked at the proof and now it makes sense. $\endgroup$ – Merry Feb 26 '14 at 18:23

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