It is always true that $\|f(t)\|_1 \geq \|f(t)\|_2 \geq \|f(t)\|_\infty$ for integral norms??

Question

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:

1. $||f||_1 \geq ||f||_2$
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: $$ \int_{-\infty}^{\infty} |f(t) \cdot g(t)| dt \leq \sqrt{\int_{-\infty}^{\infty} |f(t)|^2 dt} \cdot \sqrt{\int_{-\infty}^{\infty} |g(t)|^2 dt} $$

Answer 1

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.

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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 $\|g\|_\infty \le \|g\|_2 \le \|g\|_1$.

Answer 2

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 $$ \int_X |f\cdot g|\,d\mu \le (\int_X |f|^p\,d\mu)^{1/p}(\int_X |g|^q\,d\mu)^{1/q},\quad \frac1p+\frac1q=1. $$

Answer 3

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 $f$, $\Vert f \Vert_2 \leq \Vert f\Vert_1$.