Normed sequence space

• Apr 13th 2010, 06:25 AM
ejgmath
Normed sequence space
Let $\displaystyle \ell^1(\mathbb{N},\mathbb{R})$ denote the space of sequences $\displaystyle x(j)$ such that $\displaystyle \sum_{j=1}^{\infty}|x(j)|<\infty$, equipped with the norm $\displaystyle \|x\|_1=\sum_{j=1}^{\infty}|x(j)|$. Let $\displaystyle \|x\|_2$ denote the usual norm $\displaystyle (\sum_{j=1}^{\infty}|x(j)|^2)^\frac{1}{2}$ on $\displaystyle \ell^2(\mathbb{N},\mathbb{R})$.
Show that $\displaystyle \|x\|_2\leq\|x\|_1$, and hence deduce that $\displaystyle \ell^1(\mathbb{N},\mathbb{R})\subseteq\ell^2(\math bb{N},\mathbb{R})$.

Any help would be appreciated. Thanks in advance
• Apr 13th 2010, 07:05 AM
Jose27
Quote:

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

Any help would be appreciated. Thanks in advance

Use the fact that since the series converges the general term $\displaystyle |x_j| \rightarrow 0$ and that for $\displaystyle |t|<1$ it is valid $\displaystyle |t|^2<|t|$
• Apr 13th 2010, 07:14 AM
Laurent
Quote:

Originally Posted by Jose27
Use the fact that since the series converges the general term $\displaystyle |x_j| \rightarrow 0$ and that for $\displaystyle |t|<1$ it is valid $\displaystyle |t|^2<|t|$

This will prove the inclusion, but not the comparison of norms, I think.

For that one, you can write $\displaystyle \left(\sum_{j=1}^n |x_j|\right)^2\geq \sum_{j=1}^n |x_j|^2$ for all $\displaystyle n$, 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 $\displaystyle n\to\infty$.
• Apr 13th 2010, 07:19 AM
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. Also, how do you deduce the second part of the question? I have included the revised version below. Thanks

Quote:

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

Any help would be appreciated. Thanks in advance

• Apr 13th 2010, 07:27 AM
Laurent
Quote:

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...

Quote:

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

(Also, Jose27 suggested a direct proof, without the norm inequality)
• Apr 14th 2010, 05:14 AM
ejgmath
Could someone help explain the logic here?

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

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

Many thanks to anyone who can help
• Apr 14th 2010, 05:43 AM
Enrique2
Just think that $\displaystyle x\in l_1$ means that $\displaystyle \|x\|_1$ is finite, and the same for $\displaystyle l_2$ with $\displaystyle \|\cdot\|_2$. You can look your geometric argument by the point of view that there are vectors with "infinity length" with the measure $\displaystyle \|\cdot\|_1$, but with finite length with the smaller masure $\displaystyle \|\cdot\|_2$, as for example $\displaystyle x=(1/n)_n$. The contrary is not possible because this inequality