Triangle Inequality my balls

Sup /sci/.

well I had this problem twice in 2 days while solving some shit on L^p spaces, I can't figure if |x+y|^p =< |x|^p + |y|^p . Seems quite good to use it like this is true, but it seems rather false when I look at an example : if x and y are positive, |x+y|^p = (x+y)^p = Newtonfag(x,y) >> x^p + y^p...

Problem is I saw this used in a demonstration of Clarkson inequality which is quite simpler than the one I know and use. It uses the fact that |x+y|^p =< |x|^p + |y|^p , |x-y|^p =< |x|^p + |y|^p and conclude. The second one may be right since the minus is there but I can't accept the first.

As I saw this shit in some über book, I might have overlooked something or he writers are just faggots. Some proof/advice/confirmation ?

Tits can't be granted but I may answer with an ASCII tits.

It's not true if you mean the absolute value of x+y to the pth power. But that's not what you actually meant, is it? What you actually meant is that the pth-norm of x+y is less than the pth norm of each individually.

In L^p spaces, this becomes (|x+y|^p)^1/p =< (|x|^p)^1/p + (|y|^p)^1/p. Try working with that and see if you can understand it generally for all p>=1. You might even try the cases of p=1 and p=2 to just make sure that it is true.

Also note, when x is n dimensional, |x|^p becomes |x_1|^p+...+|x_n|^p.

>>2
Actually, it was what I meant, but it's not the same if I consider point or function, and I was doing this interpretation, which is bad. So with function, in this case, we have |f+g|^p =< |f|^p + |g|^p. Thing is, what exactly is |f| ??

It's certainly not Np(f)^(1/p) (where Np=|.|_p to avoid further confusions) because the writer used distinct notations for the absolute value (the usual |.|) and the pth-norm (the usual double bar and p-indexed). He simply wrote "|f+g|^p =< |f|^p + |g|^p" which has no sense if f and g are looked at like functions (what could be the absolute value of a non evaluated function) ?? Moreover, Np(f) is defined by an integral, you can't just put the exposant wherever you want.

I guess the author wanted to wrote "for µ-nearly all x in our mesurable space, |f(x)+g(x)|^p =<|f(x)|^p + |g(x)|^p" and did'nt write x to lighten the notations. But then it's like the first case where we were considering reals numbers x and y, so it's false. I can't think but that he failed badly.

The teacher gave us a correct formula for majoring |a+b|^p today, and it's quite easy to demonstrate btw : |a+b|^p =< 2^(p-1)*(|a|^p+|b|^p).

I consider the aforementionned demonstration of Clarkson Idendity as being bullshit and I will stick to the former one I know.

[math]
\sum_{n=0}^\infty \frac{1}{(2n+1) n!}
[\math]

$$\sum_{n=0}^\infty \frac{1}{(2n+1) n!}$$

[latex]
\sum_{n=0}^\infty \frac{1}{(2n+1) n!}
[\latex]

\sum_{n=0}^\infty \frac{1}{(2n+1) n!}