How to prove $L^{\infty}(\mathbb R) \cap L^{1} (\mathbb R) \subset L^{2}(\mathbb R)$

Question

How to prove $$L^{\infty}(\mathbb R) \cap L^{1} (\mathbb R) \subset L^{2}(\mathbb R),$$

$\mathbb R$ being the real value domain. Why is it that having the function upper-bounded and lower bounded with integrability allows the square to be integrable. L of course are Lp norms for functions.

Answer 1

For $f\in L^1$ we have $\int|f|<\infty$.

$$f\in L^\infty\cap L^1\implies \int|f|^2=\int|f|\cdot |f|\leq \|f\|_\infty\int|f|<\infty\implies f\in L^2.$$

Note that for any measurable $f$ we define $\|f\|_\infty=\inf\big\{M:\lambda(x:|f(x)|>M)=0\big\}$. Now, from definition of infimum we have $\lambda\big(\{x:|f(x)|>\|f\|_\infty+1/n\}\big)=0$ for all $n\in\Bbb N$. Therefore, $$ \begin{split} \lambda\big(\{x:|f(x)|>\|f\|_\infty\}\big)&=\lambda\bigg(\bigcup_{n\in\Bbb N }\big\{x:|f(x)|>\|f\|_\infty+1/n\big\}\bigg)=0\\ \\ \implies &|f|\leq \|f\|_\infty\text{ a.e. }\lambda. \end{split} $$

Answer 2

Actually, the property you are pointing out is fairly more general. In fact, for any $1\leq p < q \leq +\infty$, if you have a function $$ f\in L^p(\mathbb{R})\cap L^q(\mathbb{R}), $$ then, $f$ belongs to all $L^r(\mathbb{R})$ for all $r\in(p,q)$. In fact, let assume that $q<+\infty$ (since the proof for $q=\infty$ is already in the other answers). Let's consider $\theta\in (0,1)$ such that $$ \dfrac{1}{r}=\dfrac{1-\theta}{p}+\dfrac{\theta}{q}. $$ Notice that such $\theta$ always exists since $r\in(p,q)$. Now, notice that with this definition we can write $$ \Vert f\Vert_{L^r(\mathbb{R})}^r=\int \vert f(x)\vert^rdx=\int \vert f(x)\vert^{(1-\theta)r}\vert f(x)\vert^{\theta r}. $$ Then, by Hölder's inequality, noticing that $1=\tfrac{(1-\theta)r}{p}+\tfrac{\theta r}{q}$, we obtain $$ \Vert f\Vert_{L^r(\mathbb{R})}^r\leq c\big\Vert \vert f\vert^{(1-\theta)r}\big\Vert_{L^{p/((1-\theta)r)}}\big\Vert \vert f\vert^{\theta r}\Vert_{L^{q/\theta r}(\mathbb{R})}=c\Vert f\Vert_{L^p}^{(1-\theta)r}\Vert f\Vert_{L^q}^{\theta r}<+\infty. $$ PS: Notice that this proof still holds in $\mathbb{R}^n$ for any dimension $n\in\mathbb{N}$.