Re: Directional derivative in l^1 space

In article
<2622413.1194958530126.JavaMail.jakarta@xxxxxxxxxxxxxx
orum.org>,
precarion <precarion@xxxxxxxx> wrote:

Hello!

Can anyone help me with this problem:
"Let:

X = l^1 := { (x_n) - sequences of real numbers, such that sum of |x_n| for n
= 1 to +oo is less then infinity};

f = ||(x_n)|| := sum of |x_n| for n = 1 to +oo;

(x_n)* - some choosen sequence from X.

Check if the following to statements are equivalent:

(1) for every sequence h in X there exists a directional derivative f'_h
(x_n)* (it means: the directional derivative of f on (x_n)* in the direction
given by vector h),
(2) every term of (x_n)* is nonequal to 0 (i.e.: for every n in N: (x_n)* !=
0 )."

------
Chris

Hint: Try it in R^2 first, where for example you see
D_h(f)(1,-1) =
h_1 - h_2. In l^1 the guess would be D_h(f)(x) =
sum(n=1,oo) a_n*h_n,
where a_n = 1 if x_n > 0, a_n = -1 if x_n < 0.

So far I have managed to prove that (1) => (2), by showing - using contraposition [NOT (2)] => [NOT (1)] - we can just take a vector/sequence h = (h_n), such that:

(1) h_(n_0) = 1, where n_0 is index of term of a sequence (x_n)*, such that x_(n_0) = 0,
(2) h_n = 0, for every other indexes n.

But I still can't prove that (2) => (1)... It seems that it has to be proved from definition of (directional) derivative, but I'm not sure how to point it out...

Thanks for your hint, anyway... :)

------
Chris
.