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The problem is to establish the following inclusions:

Let $1 \leq p < q \leq \infty$. Then $l_p \subset l_q$, but $L_p[a,b] \supset L_q[a,b]$ as well as $L_p (\mathbb{T}) \supset L_q(\mathbb{T})$ and $L_p(R)$, $L_q(R)$ are not contained one in another.

My thoughts about the first:

If the sequence from $l_q$ was containing just of positive numbers, we would have the following: $x \in l_p =>$ starting with some $N$ holds $x^p_n + x^p_{n+1}+... < \epsilon$ which is possible only if $x_n, x_{n+1},... <1$. In this case surely $x^q_n + x^q_{n+1} + ... < \epsilon$ as well. But the sequence mustn't be containing of positive numbers, so my argument doesn't work.

Regarding the rest part of exercise, probably I have maneged to don't know some needed property of Lebesgue integral.

mathreadler
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Invincible
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1 Answers1

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Assuming $x=(x_i)\in l^p$, then for arbitrary $\epsilon>0$, $\exists k\in \mathbb{N}$ such that $\sum_{i=k}^{\infty}|x_j|^p<\epsilon$. So for $j\geq k$, $|x_j|<\epsilon^{1/p}$. As $|x_j|^q=|x_j|^{q-p}|x_j|^p$ and $q-p>0$. Then $\sum_{j=k}^{\infty}|x_j|^q=\epsilon ^{(q-p)/p}\sum_{j=k}^{\infty}|x_j|^p$, which is can be made arbitrarily small. So $x\in l^q$.

To show $L_q[a,b]\subset L_p[a,b]$ for $1\leq p<q<\infty$, let us define $A=\{x\in [a,b]: |f(x)|>1\}$, then $\mu(A^c)\leq b-a$. Now $\int_{[a,b]}|f|^p d\mu=\int_A|f|^p d\mu+\int_{A^c}|f|^p d\mu \leq \int_A|f|^q d\mu +(b-a)$. So if $f\in L_q[a,b]$ then $f\in L_p[a,b]$.

Note: For $\mathbb{R}$ we can not use this argument since $\mu(\mathbb{R})$ is not finite.

Happy
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  • Oh, I somehow forgot that we are summing the absolute values of $x_j$, not just $x_j$! Thanks. – Invincible Apr 19 '17 at 14:27
  • You are welcome. If you think your problem is solved, you might accept my answer. – Happy Apr 19 '17 at 14:59
  • @ Happy, I've accepted the answer but just a little question remains. For infinite measures this way doesn't work, but it doesn't follow from this that there is no inclusion $L_p \subset L_q$ (or conversly). Could you give me a hint about that? – Invincible Apr 19 '17 at 16:44
  • You can use the fact $\int_{0}^{1}\frac{1}{x}dx$ and $\int_{1}^{\infty}\frac{1}{x}dx$ are not finite. – Happy Apr 19 '17 at 17:01