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If $f \in L_p \cap L_q$ and $1 \leq p < r < q < \infty$ then $f \in L_r$.

Proof: Let $E = \{ x \in X: \left\lvert f \right\rvert \leq 1 \}$. Then, $$\int \left\lvert f \right\rvert^r \mathrm{d}\mu = \int_E \left\lvert f \right\rvert^r \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^r \mathrm{d}\mu \leq \int_E \left\lvert f \right\rvert^p \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^q \mathrm{d}\mu$$ $$\leq \int \left\lvert f \right\rvert^p \mathrm{d}\mu+\int \left\lvert f \right\rvert^q \mathrm{d}\mu < \infty.$$ Thus, $f \in L_r$.

I just want to make sure I am somehow not missing anything. Thanks

2 Answers2

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Your proof is correct. A shortcut is as follows: for every real number $t$, $|t|^r\leqslant|t|^p+|t|^q$ hence, for every $f$ in $L^p\cap L^q$, $\|f\|_r^r\leqslant\|f\|_p^p+\|f\|_q^q$ is finite, that is, $f$ is in $L^r$.

Did
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  • Can you tell me the name of the innequality or how to prove it ?, and if p < r why bother with it in the innequality if all you want to do is prove that $|t|^r$ is finite ? – Michael T Mckeon Dec 17 '13 at 06:22
  • @MichaelTMckeon No name I would be aware of. The part $|t|^p$ of the upper bound is needed for the case when $|t|\lt1$. – Did Dec 17 '13 at 08:32
  • I managed to figure that out as well, and the innequality wasnt to difficult to prove once I started to think about it. But thanks for your reply and suggestion above. – Michael T Mckeon Dec 18 '13 at 05:50
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The following step is wrong (thank for the comment point to me, this is a clever step!) $$ \int_E \left\lvert f \right\rvert^r \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^r \mathrm{d}\mu \leq \int_E \left\lvert f \right\rvert^p \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^q \mathrm{d}\mu. $$ Alternatively, my thought is to prove the following $$ \|f\|_p^m \|f\|_q^n \ge \|f\|_r^{m+n}\quad \text{for $r=\frac{pm+qn}{m+n}$}. $$

Ma Ming
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    Instead of just claiming that a step is wrong, you might want to say why.. In fact, for $0 \leq x \leq 1$ we have $x^p \geq x^r$ and for $x \geq 1$ we have $x^r \leq x^q$ which justifies the "wrong" step by the choice of $E$. – Martin May 09 '13 at 09:00
  • @Martin Thanks! It is a nice proof. – Ma Ming May 09 '13 at 09:03
  • So the proposed solution is correct. +1 for giving Lyapunov's interpolation inequality as an alternative approach. – Martin May 09 '13 at 09:11
  • The interpolation inequalities at the end of the post build on the inequality proposed in the question, hence it seems odd to invoke them to prove it. – Did May 09 '13 at 09:47
  • @Did You can prove that by Young's inequality. – Ma Ming May 09 '13 at 09:51
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    I know, thank you, and this fact is off-topic: Young's inequality is more sophisticated than the inequality $a^r\leqslant a^p+a^q$ (and more work is required to adapt it to the case asked here). – Did May 09 '13 at 09:57