||L||, transformations

1. Sep 18, 2008

j0k3R_

1. The problem statement, all variables and given/known data

for L, M: V -> W, L, M, linear let||L|| = sup{|L(v)|: v in V, |v| <= 1}

show ||L + M|| < ||L|| + ||M||

2. Relevant equations

3. The attempt at a solution

so is it true that if |L(x) + M(x)| defines a sup for L + M (x for which |L(x) + M(x)| is the sup), then it also defines a sup for L and sup for M, as L and M are both defined on V? im getting caught up on this and im thinking maybe theres a simpler way, i.e., either defining sets and using the standard results for sup on sets, or to exploit the linearity in a clever way

think im overlooking something simple

2. Sep 18, 2008

Pick a vector $$\mathbf v$$ such that $$\|\mathbf v \| \le 1$$.
Then what can you do with this?

$$|L(\mathbf v) + M(\mathbf v)| \le |L(\mathbf v)| + |M(\mathbf v)|$$

(think about the given norm and the r.h.s. first, then think again about the definition of this norm and the l.h.s. remember that $$\mathbf v$$ is arbitrary )

3. Sep 18, 2008

j0k3R_

suppose we have v satisfying the conditions. then |L(v) + M(v)| <= |L(v)| + |M(v)| <= ||L|| + ||M|| so sup(|L(v) + M(v)|) = ||L + M|| <= ||L|| + ||M||.

4. Sep 18, 2008

BINGO!

One minor quibble - you would usually see your final line written in this order:

$$||L + M || = \sup_{||\mathbf v || =1}|(L+M)\mathbf v| \le ||L|| + ||M||$$

5. Sep 18, 2008

j0k3R_

i got mthis solution in few second but am spending several hours thinking about it

but i do not have confidence i am thinking "this is silly something i did wrong, this norm can be complex structure etc" thinking about something extra etc.etc

okay thanks for this very much :)

oof