Normed Space $L^2$ and Vector Space $l^2$ I am really confused about $L^2$-Space and $l^2$-Space. Can anybody explain what is the difference between the $L^p$ and $l^p$ space?
Also in $L^p$-Spaces, we see that $L^2 \subset L^1$ (for $L^2[(0,T_f)], T_f >0 $). On the other hand for $l^p$-Spaces, we see that $l^1 \subseteq l^2$.
How can we show that $l^1 \subseteq l^2$?
I don't understand why(/how) does the subset sign change for the two different cases. Thanks in advance.
 A: Let $\mu$ be a measure on some measurable space $X$. You may then define:
$$L^p(X,\mu):= \{f:X\to\Bbb C\,\mid f\text{ measurable and }\int_X|f|^pd\mu<\infty\}$$
For example if $X=\Bbb N$ and $\mu_c$ is the counting measure then a function in $L^p(\Bbb N, \mu_c)$ is just a sequence $f(n)$ satisfying $\sum_{n\in\Bbb N} |f(n)|^p<\infty$, ie this is $\ell^p(\Bbb N)$. Similarly if $X$ is some interval in $\Bbb R$ and $\mu$ is the Lebesgue measure you recover the usual $L^p$ space.
With this perspective whether or not you get some kind of an inclusion $L^p(X,\mu)\subseteq L^q(X,\mu)$ is a property of the measure space $(X,\mu)$.

Definition (not standard names)
Call a measure space $(X,\mu)$ IR-divergent if $\mu(X)=\infty$, call $(X,\mu)$ UV-divergent if for any $\epsilon$ there is a measurable subset $A\subseteq X$ with $0<\mu(A)<\epsilon$.

Clearly $\Bbb N$ with the counting measure is IR-divergent but not UV-divergent while $[a,b]$ with the Lebesgue measure is UV-divergent but not IR-divergent.

Lemma
If $(X,\mu)$ is not IR-divergent then for any $p≤q\in[1,\infty)$ you've got $L^p(X,\mu)\subseteq L^q(X,\mu)$.

Proof:
Let $f\in L^q(X,\mu)$. Then
$$M(x):= \max(1, |f(x)|^q)$$
is integrable since $|f(x)|^q$ is integrable and since the constant function $1$ is integrable. Further $M(x)$ clearly majorises $|f|^p$ since $q≥p$ - hence $|f|^p$ is integable giving $L^p(X,\mu)\subseteq L^q(X,\mu)$.

Lemma
If $(X,\mu)$ is not UV-divergent then for any $p≤q\in[1,\infty)$ you've got $L^p(X,\mu)\supseteq L^q(X,\mu)$.

Proof:
If $f\in L^p(x)$ then $A:=|f(x)|^{-1}([1,\infty))$ has finite measure so by $X$ not being UV-divergent there exists a finite covering $A= \bigcup_{i=1}^n A_i$ where none of the $A_i$ admit sub-sets of smaller (non-zero) measure. In particular $|f(x)|$ must be constant almost everywhere on the individual $A_i$ as such $|f(x)|$ must admit an essential supremum on $A$, call this superemum $S$.
$$M(x):=\begin{cases} S^q & x\in A\\ |f(x)|^p& x\notin A\end{cases}$$
is integrable since $A$ has finite measure and $|f(x)|^p$ is integrable. Further it majorises $|f(x)|^q$ since outside of $A$ you have $|f(x)|<1$ whence $|f(x)|^q ≤ |f(x)|^p$ and inside of $A$ you have $|f(x)|^q ≤ S^q$. It follows that $L^p(X,\mu)\supseteq L^q(X,\mu)$.

So the thing you have seen is an expression of the fact that $\Bbb N$ does not admit subsets of arbitrarily small measure and that $[a,b]$ has bounded Lebesgue measure.
