# Sequence spaces are subsets for p<q

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 $l^p\subset l^q$

• What have you tried? How are $|x|^p$ and $|x|^q$ related when $|x| < 1$? Why is this the only case that matters? – John Hughes Feb 14 '17 at 19:25

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 $\| x \|_q \leq \| x \|_p$.

• Could you please elaborate summing over n part.....I tried summing but i couldn't arrive at the same solution as yours. @NeedForHelp – Accidental Genius Feb 15 '17 at 19:53
• @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 $$\|x\|_q \leq\| x\|_p$$ – NeedForHelp Feb 15 '17 at 20:08
• Wow....yes thank you @NeedForHelp So if $||x||_q <= ||x||_p$ how is $l^p \subset l^q$ ? – Accidental Genius Feb 15 '17 at 20:11
• @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. – NeedForHelp Feb 15 '17 at 20:15
• You saved me atleast 5 points in the exam to come. Thanks a ton – Accidental Genius Feb 15 '17 at 20:18

Let $1\leq p<q<+\infty$. If $x\in l_p$, then $\sum_n |x_n|^p<+\infty$, so $|x_n|^p\to 0$, and that means $|x_n|\to 0$. So $x$ is a bounded sequence, $|x_n|<M$, (so we have proven that $x\in l_\infty$).

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

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