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

functional-analysisfunctionsnormed-spacesreal-analysis

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

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