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:

$$ \int_{-\infty}^{\infty} |f(t) \cdot g(t)| dt \leq \sqrt{\int_{-\infty}^{\infty} |f(t)|^2 dt} \cdot \sqrt{\int_{-\infty}^{\infty} |g(t)|^2 dt} $$

## Best Answer

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