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Sorry if this is very basic but here's a question.

Let $\mathbf{v}=(v_1,\ldots, v_n)\in k^n$ where $k=\bar{k}$.

Why do we have
$$ \sqrt{\sum_{i=1}^n |v_i|^2} \leq \sum_{i=1}^n |v_i|, $$ where the left-hand side can be thought of as the $2$-norm $\|\mathbf{v}\|_2$ on $L^2(k^n)$?

$\mathbf{General \; case}$: If this is true for $p=2$-norm, I am guessing that this is true for all $p\geq 1$: $$ \left( \sum_{i=1}^n |v_i|^p\right)^{1/p}\leq \sum_{i=1}^n |v_i|. $$

In fact, does this inequality hold when $p$ is a rational number?

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up vote 4 down vote accepted

Square both sides to get: $$ \sum_{i=1}^n |v_i|^2 \leq \left(\sum_{i=1}^n |v_i|\right)^2 $$

Expand the RHS using the multinomial theorem to see that it's equal to the LHS plus a number of non-negative terms. Hence the inequality holds.

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+1, but tiny nitpick: some of the extra terms could be 0, so you should use "non-negative" instead of "positive" – Zev Chonoles Jun 29 '12 at 8:56
@ZevChonoles - Good point. Fixed now. Thanks for the comment! – Ayman Hourieh Jun 29 '12 at 8:57
Thanks Ayman! I had forgotten such tricks! =) – math-visitor Jun 29 '12 at 8:59

Or geometrically, the diagonal distance between opposite corners in an n-dimensional cuboid is smaller, than walking along the edges.

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Thanks Paxinum for the geometric visualization! – math-visitor Jun 29 '12 at 9:16

To answer your second question, yes the inequality holds for all rational $p\geq1$. In fact, the inequality holds for all $p\geq1.$

To see this note that the function $x\mapsto x^{p}$ is convex, so that if $x,y\geq0$, then

$$\frac{(x+y)^p-x^p}{(x+y)-x}\geq \frac{y^p-0^p}{y-0}\Rightarrow (x+y)^p\geq x^p+y^p.$$

By induction, if $x_i\geq0$,

$$(x_1+\cdots+x_n)^p\geq x_1^p+\cdots+x_n^p.$$

This is equivalent to your second assertion.

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Nice, and thanks! So a similar argument can be applied to a family of convex functions! – math-visitor Jun 29 '12 at 10:56
@math-visitor: Yes, although in general you'll have an $nf(0)$ floating around. – Benji Jun 29 '12 at 15:48

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