1. MSE link The case of $m(E)=0$ or $\|f\|_{_{L^{\infty}(E)}}=0$ is trivial. When $\displaystyle m(E)>0$ and $\displaystyle \|f\|_{L^{\infty}(E)}>0$, assuming $\displaystyle c$ satisfying $\displaystyle 0<c<\|f\|_{L^{\infty}(E)}$, there exists $\displaystyle E_{c}\subset E$, which is measurable with $\displaystyle m(E_{c})>0$ and satisfies $\displaystyle |f|>c$ on $\displaystyle E_{c}$. Then we have $\displaystyle m^{\frac{1}{p}}(E_{c})\le\left(\int_{E_{c}}|f|^{p}\right)^{\frac{1}{p}}\le\left(\int_{E}|f|^{p}\right)^{\frac{1}{p}}\le m^{\frac{1}p}(E)\|f\|_{L^{\infty}(E)}$. Now let $\displaystyle p\to\infty$, we know $\displaystyle c\le \lim_{p\to\infty}\|f\|_{_{L^{p}(E)}}\le\|f\|_{L^{\infty}(E)}$, that is $\displaystyle \lim_{p\to \infty}\|f\|_{_{L^{p}(E)}}=\|f\|_{_{L^{\infty}(E)}}$, by the arbitrariness of $c$.

2. MSE link Using Holder inequality, we can get this conclusion by $\displaystyle \int_{E}\left|f(x)\right|^{q}\mathrm{d}x\le\left(\int_{E}\left|f(x)\right|^{q\cdot \frac{p}{q}}\mathrm{d}x\right)^{\frac{q}{p}}\left(\int_{E}1^{\frac{p}{p-q}}\mathrm{d}x\right)^{\frac{p-q}{p}}=\left(\int_{E}\left|f(x)\right|^{p}\mathrm{d}x\right)^{\frac{q}{p}}\left(m(E)\right)^{\frac{p-q}{p}}<\infty$ when $\displaystyle 0<q\le p\le \infty$, immediately.

3. For the case of $\displaystyle 1\le q<p<\infty$, let $\displaystyle f(x)=x^{-\frac{1}{p}}\mathbf{1}_{0<x\le1}$, we have $\displaystyle f\in L^{q}$ but $\displaystyle f\notin L^{p}$, so $\displaystyle L^{p}\not\subset L^{q}$. Then let $\displaystyle g(x)=(1+x)^{-\frac{1}{q}}$, we have $\displaystyle g\in L^{p}$ but $\displaystyle g\notin L^{q}$, so $\displaystyle L^{q}\not\subset L^{p}$. For the case of $\displaystyle p=\infty,q<\infty$, let $\displaystyle h(x)=x^{-\frac{1}{2q}}\mathbf{1}_{0<x\le1}$, we have $\displaystyle h\in L^{q},h\notin L^{\infty}$ and $\displaystyle \mathbf{1}_{\mathbb{R}}\in L^{\infty},\mathbf{1}_{\mathbb{R}}\notin L^{q}$, so $\displaystyle L^{p}\not\subset L^{q}$ and $\displaystyle L^{q}\not\subset L^{p}$.

4. From Real Analysis by Folland If $f \in L^p$, let $E = \{x : |f(x)| > 1\}$ and set $g = f \chi_E$ and $h = f \chi_{E^c}$.
   Then $|g|^r = |f|^r \chi_E \leq |f|^q \chi_E$, so $g \in L^r$, and $|h|^s = |f|^s \chi_{E^c} \leq |f|^q \chi_{E^c}$, so $h \in L^s$.
   (For $s = \infty$, obviously $\|h\|_\infty \leq 1$.)

5. By Holder inequality, we know $\displaystyle \|fg\|^{p}_{L^{p}}=\int_{\mathbb{R}}\left|f(x)g(x)\right|^{p}\mathrm{d}x\le\left(\int_{\mathbb{R}}\left|f(x)\right|^{q}\mathrm{d}x\right)^{\frac{p}{q}}\left(\int_{\mathbb{R}}|g(x)|^{r}\mathrm{d}x\right)^{\frac{p}{r}}=\left\|f\right\|_{L^{q}}^{p}\left\|g\right\|_{L^{r}}^{p}$, that is, $\displaystyle \|fg\|_{L^{p}}\le\|f\|_{L^{q}}\|g\|_{L^{r}}$.

6. From Classical Fourier Analysis by Loukas Grafakos Let $g(x)=|f(x)| x^{1-\frac{q}{p}} \in L^p\left(\mathbb{R}^{+},\frac{dx}{x}\right)$ and $h(x)=x^{-\frac{q}{p}} \chi_{[1, \infty)} \in L^1\left(\mathbb{R}^{+},\frac{dx}{x}\right)$. Now compute
   $$(g * h)(x)=\int_0^{\infty}|f(t)| t^{1-\frac{q}{p}}\left(\frac{x}{t}\right)^{-\frac{q}{p}} \chi_{[1, \infty)}\left(\frac{x}{t}\right) \frac{d t}{t}=x^{-\frac{q}{p}} \int_0^x|f(t)| d t$$since $\frac{x}{t} \in[1, \infty)$ if and only if $t \in(0, x]$. We have
   $$\|g * h\|_{L^p}=\left(\int_0^{\infty}\left(\int_0^x|f(t)| d t\right)^p x^{-q} \frac{d x}{x}\right)^{\frac{1}{p}}=\left(\int_0^{\infty}\left(\int_0^x|f(t)| d t\right)^p x^{-q-1} d x\right)^{\frac{1}{p}}$$
   also
   $$\|g\|_{L^p}=\left(\int_0^{\infty}|f(t)|^p t^{p-q-1} d t\right)$$
   and
   $$\|h\|_{L^1}=\int_0^{\infty} t^{-\frac{q}{p}} \chi_{[1, \infty)} \frac{d t}{t}=\int_1^{\infty} t^{-\frac{q}{p}-1} d t=\frac{p}{q}$$
   The desired inequality now follows from Minkowski's inequality.

7. From Real Analysis by Folland If $s=\infty$, we have $|f|^q \leq\|f\|_{\infty}^{q-p}|f|^p$ and $\theta=p / q$, so
   $$\|f\|_q \leq\|f\|_p^{p / q}\|f\|_{\infty}^{1-(p / q)}=\|f\|_p^\theta\|f\|_{\infty}^{1-\theta} .$$
   If $s<\infty$, we use Holder's inequality, taking the pair of conjugate exponents to be $p / \theta q$ and $s /(1-\theta) q$ :
   $$\begin{aligned} \int|f|^q & =\int|f|^{\theta q}|f|^{(1-\theta) q} \leq\left\||f|^{\theta q}\right\|_{p / \theta q}\left\||f|^{(1-\theta) q}\right\|_{s /(1-\theta) q} \\ & =\left[\int|f|^p\right]^{\theta q / p}\left[\int|f|^s\right]^{(1-\theta) q / s}=\|f\|_p^{\theta q}\|f\|_s^{(1-\theta) q} \end{aligned}$$
   Taking $q$ th roots, we are done.

8. From Real Analysis by Folland First, if $g \in C_c$, for $|y| \leq 1$ the functions $\tau_h g$ are all supported in a common compact set $K$, so
   $$\int\left|\tau_h g-g\right|^p \leq\left\|\tau_h g-g\right\|_\infty^p m(K) \rightarrow 0 \text { as } y \rightarrow 0$$
   Now suppose $f \in L^p$. If $\epsilon>0$, by the density of $C_{c}$ function, there exists $g \in C_c$ with $\|g-f\|_p<\epsilon / 3$, so
   $$\left\|\tau_h f-f\right\|_p \leq\left\|\tau_h(f-g)\right\|_p+\left\|\tau_h g-g\right\|_p+\|g-f\|_p<\frac{2}{3} \epsilon+\left\|\tau_h g-g\right\|_p,$$
   and $\left\|\tau_h g-g\right\|_p<\epsilon / 3$ if $y$ is sufficiently small.

9. MSE link Notice that $\displaystyle L^{p}([0,1])\subset L^{1}([0,1])$ for $\displaystyle 1\le p\le \infty$, then $\displaystyle f\in L^{1}([0,1])$. We can know $\displaystyle \int_{0}^{1}p(t)f(t)\mathrm{d}t=0$ for any polynomial $\displaystyle p(t)$ from the condition. Notice that $\displaystyle e^{-x^{2}}$ can be regard as a polynomial because of Taylor's formula, we have $\displaystyle f_{\delta}=\frac{1}{\sqrt{\pi}\delta}\int_{0}^{1}f(x)e^{-\frac{(x-y)^{2}}{\delta^{2}}}\mathrm{d}y=0$. And because of $\displaystyle \|f_{\delta}-f\|_{L^{1}}\to0$ when $\displaystyle \delta\to0$, we have $\displaystyle f\equiv0$.

10. MSE link Consider a sequence $\displaystyle \{f_{z}\}\in L^{\infty}([0,1]),f_{z}(x)=\begin{cases}0&x\in[0,z]\\1&x\in(z,1]\end{cases},z\in[0,1]$, we know $\displaystyle \forall m\neq n,\|f_{m}-f_{n}\|_{L^{\infty}}=1$. So for any $\displaystyle g\in L^{\infty}$, there is at most only one $\displaystyle f_{a}$ satisfying $\displaystyle \|g-f_{a}\|_{L^{\infty}}<\frac{1}{2}$. So for any countable set $\displaystyle \{g_{k}\}\in L^{\infty}$, there exists $\displaystyle f\in \{f_{z}\}$ satisfying $\displaystyle \|f-g_{k}\|_{L^{\infty}}\ge \frac{1}{2},\forall k$. Then $\displaystyle \{g_{k}\}$ won't exist countable dense subset of $\displaystyle L^{\infty}$. Therefore, $\displaystyle L^{\infty}$ isn't separable.