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Possible Duplicate:
Inequality between $\ell^p$-norms

First, I'll start by simply showing you the problem:

Let $1 \leq p \leq q < \infty$. Determine that $l^p \subseteq l^p$ by proving $\Vert x \Vert_q \leq \Vert x \Vert_p$ $\forall x \in l^p$. Hint: Start with $\Vert x \Vert_p = 1$

So, well I started with $\Vert x \Vert_p = 1$ and at this point is is relatively easy to show that the inequality holds for this case. Now I need to expand it to arbitrary values. My idea was to use a sequence $x + \delta$ with $x$ being the sequence with norm $1$. This didn't work out for me because I would have needed $\Vert x \Vert_q \leq \Vert x \Vert_p$ to show it. Makes obviously no sense if you need the inequality you want to prove to prove it.

I also messed around with a few formulas, such as the triangle inequality to split it up, didn't help. My last idea was to use induction so show that if the inequality holds for a sequence $(x_1, x_2, ... , x_n, 0, 0, 0, ...)$ it also holds for $(x_1, x_2, ... , x_n, x_{n+1}, 0, 0, ...)$. Didn't work out as well.

So any ideas are appreciated, I will embrace any idea that will show me the right direction, not necessarily the whole answer .)

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    For $\Vert x\Vert_p \ne 0$, just set $y={x\over \Vert x\Vert_p}$ and apply your inequality for the norm one case. Then factor out the constant $1\over\Vert x \Vert_p$.2011-11-24

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