It is always true that $\|f(t)\|_1 \geq \|f(t)\|_2 \geq \|f(t)\|_\infty$ for integral norms?? I found on Wikipedia page for "$L_p$ Spaces" [1] that if the norm is defined as:
$$\|f(t)\|_p = \left(\int_{-\infty}^{\infty} |f(t)|^p dt \right)^\frac{1}{p}$$
with $|\cdot|$ the "absolute value", then the following relations are true:

*

*$||f||_1 \geq ||f||_2$

*$||f||_{p+a} \leq ||f||_p$ for any $p \geq 1$ and $a \geq 0$.

There is also defined that (if I made no mistake):
$$\|f(t)\|_\infty = \lim_{p \to \infty}\|f\|_p = \sup_t\{|f(t)|\}$$
So following point 2, then also I have that: $\|f\|_\infty \leq \|f\|_2$, (since $2+\infty \gg 2$).
A) I want to know if this relation $\|f(t)\|_1 \geq \|f(t)\|_2 \geq \|f(t)\|_\infty$ is always true? If not, what is needed to make it true? And also, if I have understood properly the definition of $\|f\|_\infty$.
I have tried some examples of real-valued one-variable Lebesgue-integrable and square-integrable functions using Wolfram-Alpha [2]:
a) $f_1(t) = \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}}$ the standard Gaussian distribution.
b) $f_2(t) = e^{-t} \cdot \theta(t)$, with $\theta(t)$ the standard step function.
$$
\begin{array}{| c | c : c : c | c| c|} \hline
f(t) & \|f\|_1 = \int_{-\infty}^\infty |f(t)|\,dt & \|f\|_2 = \sqrt{\int_{-\infty}^\infty |f(t)|^2\,dt} & \|f\|_\infty = \sup_t\{|f(t)|\} & ¿ \|f\|_2 \leq \|f\|_1 ? & ¿ \|f\|_\infty \leq \|f\|_2 ? \\ \hline
f_1 & 1 & \sqrt{\frac{1}{2 \sqrt{\pi}}} = 0.531 & \frac{1}{\sqrt{2\pi}} = 0.39 & \color{green}{\checkmark} & \color{green}{\checkmark} \\ \hdashline
f_2 & 1 & \sqrt{\frac{1}{2}} = 0.707 & 1 & \color{green}{\checkmark} & \color{red}{\times}\\ \hline
\end{array}
$$
Since $f_2(t)$ doesn't fulfill the relation from Wikipedia, What I am doing wrong?
B) Is $\int_{-\infty}^\infty |f(t)|\,dt \leq \sqrt{\int_{-\infty}^\infty |f(t)|^2\,dt}$?? or $\sqrt{\int_{-\infty}^\infty |f(t)|^2\,dt} \leq \int_{-\infty}^\infty |f(t)|\,dt$?? or Nothing can be stated about it beforehand knowing $f(t)$??
C) Is $ \sup_t\{|f(t)|\} \leq \sqrt{\int_{-\infty}^\infty |f(t)|^2\,dt}$?? or $\sqrt{\int_{-\infty}^\infty |f(t)|^2\,dt} \leq \sup_t\{|f(t)|\}$?? or Nothing can be stated about it beforehand knowing $f(t)$??
D) Is the Hölder's inequality always valid for any $f(t)$ if the integration domain is $(-\infty\,;\,+\infty)$???
On Wikipedia [3] says the following related inequality is true for any $p \leq 1$ and $\frac{1}{p}+\frac{1}{q}=1$ for some integration interval $S$, and I want to know if is valid for $S=(-\infty\,;\,+\infty)$ and $p=q=2$, or equivalently:
$$ \int_{-\infty}^{\infty} |f(t) \cdot g(t)| dt \leq \sqrt{\int_{-\infty}^{\infty} |f(t)|^2 dt} \cdot \sqrt{\int_{-\infty}^{\infty} |g(t)|^2 dt} $$
 A: Your space here, Lebesgue measure on the real line, is in the "neither condition" case described on the wiki page.  These inequalities do not hold in general for this case.

For this case:
$$
\|f\|_p = \left(\int_0^1 |f(t)|^p dt\right)^{1/p} ,
$$
it is true that $\|f\|_1 \le \|f\|_2 \le \|f\|_\infty$.
For this case:
$$
\|g\|_p = \left(\sum_{k=1}^\infty |g(k)|^p\right)^{1/p} ,
$$
it is true that $\|g\|_\infty \le \|g\|_2 \le \|g\|_1$.
A: If you are considering the sequence spaces $\ell^p$, then the inequality:
$$ (\sum_i |a_i|^p)^{1/p} \le (\sum_i|a_i|^q)^{1/q},\quad p\ge q$$
holds. This could be rewritten using the norms as
$$ \|\{a_i\}\|_p \le \|\{a_i\}\|_q,\quad p\ge q.$$
This has nothing to do with functions on $\mathbb R$, and it certainly doesn't imply
$$
\text{(wrong}\rightarrow)\qquad(\int_{-\infty}^\infty |f(t)|^p\,dt)^{1/p}\le (\int_{-\infty}^\infty|f(t)|^q\,dt)^{1/q},\quad p\ge q.
$$
This is the flip of what happens in probability spaces—see my response to B).
A) $\|f\|_\infty$ is not defined as $\lim_{p\to\infty}\|f\|_p$, and this limit does not always hold (one of the reasons it is inappropriate for a definition). For a measure space $(X,\mu)$, $\|f\|_\infty = \inf\{C > 0: |f|\le C\ \text{$\mu$-a.e.}\}$. For the case of $\mathbb R$, this definition becomes
$$
\|f\|_\infty = \inf\{C>0:|f(t)|\le C\ \text{for Lebesgue-almost every $t$}\}.
$$
It is true that $\|f\|_\infty = \lim_{p\to\infty}\|f\|_p$ if $\mu(X)<\infty$, as you can see here.
B) Neither inequality holds in general, and in particular, neither inequality holds for functions on $\mathbb R$. If $\mu(X) = 1$, then $\|f\|_p\le\|f\|_q$ if $p \le q$.
C) Again, neither inequality holds in general, nor for functions on $\mathbb R$.
D) Hölder's inequality is valid for any measure space. We always have
$$
\int_X |f\cdot g|\,d\mu \le (\int_X |f|^p\,d\mu)^{1/p}(\int_X |g|^q\,d\mu)^{1/q},\quad \frac1p+\frac1q=1.
$$
A: Let $a \in \mathbb{R}_+$, and consider $f_a$ the map that sends $t$ to $1$ if $\vert t \vert \leq a/2$, and to $0$ otherwise.
Then $\int^\infty_{-\infty} f_a(t) dt = a$, so for every $p \in [1,\infty)$, $\Vert f_a \Vert_p = a^{\frac{1}{p}}$.
Let's take $a := \frac{1}{2}$. Then $\Vert f_a \Vert_1 = \frac{1}{2} < \frac{\sqrt{2}}{2} = \sqrt{\frac{1}{2}} = \left(\frac{1}{2}\right)^{\frac{1}{2}} = \Vert f_a \Vert_2$.
So, it is not true that for every $f$, $\Vert f \Vert_2 \leq \Vert f\Vert_1$.
