I am new to $ L^{p} $-spaces and am trying to prove a few things about them. Therefore, I would like to ask you whether I have gotten the following right.
Prove that $ {L^{\infty}}(I) \subseteq {L^{p}}(I) $ for all $ p \in (0,\infty) $, where $ I = [a,b] $ is a closed bounded interval, by showing that $$ \text{$ \forall $ measurable functions $ f $ defined on $ [a,b] $}: \quad \| f \|_{L^{p}} \le (b - a)^{\frac{1}{p}} \| f \|_{L^{\infty}}. $$ My idea was to invoke the fact that $$ \| f \|_{L^{p}} \stackrel{\text{def}}{=} \left( \int_{I} |f|^{p} \, d{\mu} \right)^{\frac{1}{p}} \le \| f \|_{\infty} \left( \int_{I} 1 \, d{\mu} \right)^{\frac{1}{p}} $$ for all measurable functions $ f $ defined on $ [a,b] $. Is this correct?
Prove that $ \displaystyle {L^{\infty}}(I) \subsetneq \bigcap_{p \in (0,\infty)} {L^{p}}(I) $. Inclusion is clear from (1), and a function that illustrates why the inclusion is strict is $ f(x) \stackrel{\text{def}}{=} \dfrac{1}{x^{x}} $ on the interval $ [0,1] $ (its $ L^{p} $-norm exists for each $ p \le \infty $, but it is not bounded).
Prove that $ {L^{p}}(\mathbb{R}) $ is not a subset of $ {L^{\infty}}(\mathbb{R}) $ and vice-versa. Take the function $ f(x) \stackrel{\text{def}}{\equiv} 3 $. It is not integrable over $ \mathbb{R} $, but it is bounded. For the reverse implication, take the function \begin{equation} f(x) \stackrel{\text{def}}{=} \left\{ \begin{array}{ll} \frac{1}{\sqrt{x}} & \text{if $ 0 \le x \le 1 $}; \\ 0 & \text{if $ x \in (- \infty,0) \cup (1,\infty) $}. \end{array} \right. \end{equation} It is integrable over $ \mathbb{R} $, but it is not bounded.
Maybe I have made some mistakes here or maybe I am missing something, so I appreciate any kind of help!!!
P.S.: I have just noticed that my example $ \dfrac{1}{x^{x}} $ does not work, since this function is actually bounded on $ [0,1] $.
Number 1 is correct, and so is 3.
Can you justify why the function you wrote in #2 works using a calculation?