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Can somebody prove the following statement for me

For $1\leq p \leq q \leq \infty$, $$l^p := \{(\,x_i)\,_{i\in N}| \sum_{n}|x_n|^p < \infty \}.$$ Show that $l^p\subset l^q$

Thanking you in advance.

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    What have you tried? How are $|x|^p $ and $|x|^q$ related when $|x| < 1$? Why is this the only case that matters? – 2017-02-14

2 Answers 2

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Let $x \in l^p$, $x \neq 0$. For each $n \geq 1$, $|x_n| \leq \| x \|_{p}$. But $y^q \leq y^p$ for all $0 \leq y \leq 1$ so $$ \left( \frac{|x_n|}{\| x \|_{p}} \right)^q \leq \left( \frac{|x_n|}{\| x \|_{p}} \right)^p $$ Summing over $n$, we obtain $\| x \|_q \leq \| x \|_p$.

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    Could you please elaborate summing over n part.....I tried summing but i couldn't arrive at the same solution as yours. @NeedForHelp – 2017-02-15
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    @Vasu Kolli Sure. From the inequality, we have $$\frac{1}{\| x\|_p^q}\sum_n|x_n|^q \leq \frac{1}{\| x\|_p^p}\sum_n|x_n|^p$$ which is nothing more than $$\frac{\|x\|_q^q}{\| x\|_p^q} \leq \underbrace{\frac{\|x\|_p^p}{\| x\|_p^p}}_{1}$$ and rearranging we get $$\|x\|_q^q \leq\| x\|_p^q$$ and finally $$\|x\|_q \leq\| x\|_p$$ – 2017-02-15
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    Wow....yes thank you @NeedForHelp So if $||x||_q <= ||x||_p$ how is $l^p \subset l^q$ ? – 2017-02-15
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    @Vasu Kolli Well, $x \in l^p$ means that $\| x \|_p < \infty$ and in that case $\| x \|_q \leq \| x \|_p < \infty$ hence $\| x \|_q < \infty$ also, that is to say $x \in l^q$ also. This shows that if $x \in l^p$ then $x \in l^q$, which is what $l^p \subset l^q$ means. – 2017-02-15
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    You saved me atleast 5 points in the exam to come. Thanks a ton – 2017-02-15
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    @Vasu Kolli No problem! The other answer is more direct though. The argument I shared is tricky in my opinion. – 2017-02-15
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    I think something is better than nothing. Instead of not attempting the question, I better write this. :) – 2017-02-15
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Let $1\leq p

Now, let's see that $x\in l_q$. We have:

$$\sum_n|x_n|^q \leq \sum_n |x_n|^p\cdot |x_n|^{q-p}\leq M^{q-p}\sum_n|x_n|^p<+\infty$$