Why p>1, $L^p$ and $H^p$ are essentially the same? the conclusion comes from Link (the first page)
$L^{p}$ is Lebesgue integral function space, $H^{p} $ is Hardy space. $p>1$
My question are:

*

*"Same" means that they are isomorphic ? Is this proved?


*No matter for the real function or complex function, they are always the same?
 A: One thing to keep in mind is that the Hardy space $H^p$ may be different from the Hardy space $H^p$. By which I mean, the same name and notation is attached to different, though  related, concepts. 
There are holomorphic Hardy spaces $H^p$: they consist of holomorphic functions with suitably bounded integral means. This is the origin of the theory. Yet, this precise space is not  what people have in mind when they say that $H^p=L^p$ for $p>1$.   Rather, the following holds: 

for these values of $p$ [$1<p<\infty$], $L^p(T)$ is the space $\operatorname{Re}H^p $ consisting of the boundary values of the real parts of functions in $H^p$ 

The quote is from  page 571, the third page of the article to which you linked. It helps to read beyond the first page in search of explanation of statements made there. 
As it happened, harmonic analysts quickly got tired of saying  that they are working with "the space of the boundary values of the real parts of functions in the Hardy space $H^p$", and shortened it to "the Hardy space $H^p$". Since their work is clearly different from what complex analysts did with holomorphic $H^p$ spaces, and usually involves higher dimensions, the context gives enough clues to the correct meaning of $H^p$. 
To answer your specific questions:


*

*"Same" does not mean merely isomorphic: it means they have the same elements. Abstract Banach space isomorphisms are not as important when we work with function spaces, in which the elements are concrete functions. For references, see the article in question, specifically page 571.

*No, as explained above. 
