1. ## Normed sequence space

Let $\ell^1(\mathbb{N},\mathbb{R})$ denote the space of sequences $x(j)$ such that $\sum_{j=1}^{\infty}|x(j)|<\infty$, equipped with the norm $\|x\|_1=\sum_{j=1}^{\infty}|x(j)|$. Let $\|x\|_2$ denote the usual norm $(\sum_{j=1}^{\infty}|x(j)|^2)^\frac{1}{2}$ on $\ell^2(\mathbb{N},\mathbb{R})$.
Show that $\|x\|_2\leq\|x\|_1$, and hence deduce that $\ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math bb{N},\mathbb{R})$.

Any help would be appreciated. Thanks in advance

2. Originally Posted by ejgmath
Let $\ell^1(\mathbb{N},\mathbb{R})$ denote the space of sequences $x(j)$ such that $\sum_{j=1}^{\infty}|x(j)|<\infty$, equipped with the norm $\|x\|_1=\sum_{j=1}^{\infty}|x_j|$. Let $\|x\|_2$ denote the usual norm $(\sum_{j=1}^{\infty}|x(j)|)^\frac{1}{2}$ on $\ell^2(\mathbb{N},\mathbb{R})$.
Show that $\|x\|_2\leq\|x\|_1$, and hence deduce that $\ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math bb{N},\mathbb{R})$.

Any help would be appreciated. Thanks in advance
Use the fact that since the series converges the general term $|x_j| \rightarrow 0$ and that for $|t|<1$ it is valid $|t|^2<|t|$

3. Originally Posted by Jose27
Use the fact that since the series converges the general term $|x_j| \rightarrow 0$ and that for $|t|<1$ it is valid $|t|^2<|t|$
This will prove the inclusion, but not the comparison of norms, I think.

For that one, you can write $\left(\sum_{j=1}^n |x_j|\right)^2\geq \sum_{j=1}^n |x_j|^2$ for all $n$, which comes from the expansion of the left-hand side (all terms of the expansion are nonnegative, and the right-hand side consists in a few of those). And let $n\to\infty$.

4. Hi, thanks for the response, I am really sorry but I made a small mistake in the original post. I missed the squared out of the 2-norm, does this change much. Also, how do you deduce the second part of the question? I have included the revised version below. Thanks

Originally Posted by ejgmath
Let $\ell^1(\mathbb{N},\mathbb{R})$ denote the space of sequences $x(j)$ such that $\sum_{j=1}^{\infty}|x(j)|<\infty$, equipped with the norm $\|x\|_1=\sum_{j=1}^{\infty}|x(j)|$. Let $\|x\|_2$ denote the usual norm $(\sum_{j=1}^{\infty}|x(j)|^2)^\frac{1}{2}$ on $\ell^2(\mathbb{N},\mathbb{R})$.
Show that $\|x\|_2\leq\|x\|_1$, and hence deduce that $\ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math bb{N},\mathbb{R})$.

Any help would be appreciated. Thanks in advance

5. Originally Posted by ejgmath
Hi, thanks for the response, I am really sorry but I made a small mistake in the original post. I missed the squared out of the 2-norm, does this change much.
Nothing; this was an obvious mistake we took into account...

Also, how do you deduce the second part of the question?
I think you can find it yourself. What is $\ell^1(\mathbb{N},\mathbb{R})$ by definition?...

(Also, Jose27 suggested a direct proof, without the norm inequality)

6. Could someone help explain the logic here?

I don't understand how $\ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math bb{N},\mathbb{R})$ follows from $\|x\|_2\leq\|x\|_1$.

If the vectors in $\ell^1$ have a greater length than the ones in $\ell^2$ how can $\ell^1$ be a subset of $\ell^2$?

Many thanks to anyone who can help

7. Just think that $x\in l_1$ means that $\|x\|_1$ is finite, and the same for $l_2$ with $\|\cdot\|_2$. You can look your geometric argument by the point of view that there are vectors with "infinity length" with the measure $\|\cdot\|_1$, but with finite length with the smaller masure $\|\cdot\|_2$, as for example $x=(1/n)_n$. The contrary is not possible because this inequality