$L^p$ spaces : $ p= 1,2,oo $

[Camillo]

For $lambda in RR $ let $ u_lambda $ be the function defined in $ [ 1,+oo)$ with the formula :

$u_lambda(x) = k^(-lambda)$ if $ x in[k,k+1) ; (k=1,2,....) $

For $p =1,2,oo $ determine $lambda$ so that $u_lambda in L^p(1,oo) $.

[Camillo]

Quite correct solutions

[elgiovo]

I really don't know whether or not my solution is correct, but I'd like to try:

$p=1$) $int_1^(oo) |u_lambda| dx=||u_lambda(x)||_1$ is the sum of the area of the infinity of rectangles whose width is $1$
and whose length is $1/(k^lambda)$. So $||u_lambda(x)||_1=sum_(k=1)^oo 1/(k^lambda)=zeta(lambda)1$.

$p=2$) The same line of reasoning gives $||u_lambda(x)||_2=sqrt(sum_(k=1)^oo 1/(k^(2lambda)))=sqrt(zeta(2lambda))1/2$.

$p=oo$) The minimum value such that $||u_lambda(x)||_(oo)= mbox(inf){C>=0:|u_lambda(x)|<=0 mbox( a.e.)}$, in other words
such that $u_lambda(x)$ keeps limited is $lambda=0$ (we obtain a series of unitary squares).
So $u_lambda(x) in L^(oo)(1,oo)$ iff $lambda>=0$.