Why is $L^{1} \cap L^{\infty}$ dense is in $L^{p}$?

Question

It is mentioned that using the interpolation inequality

$$\Vert f \Vert_{p} \leq \Vert f \Vert^{1/p}_{1} \Vert f \Vert_{\infty}^{1-1/p}$$

one can deduce that the space $L^{1} \cap L^{\infty}$ is dense in $L^{p}$. Does anybody knows the trick behind this? Thanks !

Answer 1

The inequality implies $L^1\cap L^\infty\subset L^p$.

Density is then clear, since the space $C_0^\infty$ of compactly supported smooth functions is dense in $L^p$ for every $\infty>p\ge 1$ and $C_0^\infty\subset L^1\cap L^\infty$.

Answer 2

Let $f\in L^p(X)$, where $X$ is a measure space, and set $$ f_n(x)=\left\{\begin{array}{lll} f(x) & \text{if} & \frac{1}{n}\le |f(x)|\le n, \\ 0 & \text{if} & |f(x)|> n \,\,\text{or}\,\,|f(x)|<\frac{1}{n}. \end{array} \right. $$ Clearly $f_n\in L^\infty(X)$ and $\|f_n-f\|_p\to 0$. The last is a consequence of Lebesgue Dominated Convergence Theorem, as $f_n(x)\to f(x)$, for every $x\in X$ and $|f_n(x)|\le |f(x)|$.

Also, the fact that $|f_n(x)|\ge \frac{1}{n}$, implies that $|f_n(x)|^{p-1}\ge \frac{1}{n^{p-1}}$ and altogether $|f_n(x)|^{p}\ge \frac{1}{n^{p-1}}|f_n(x)|$. Thus $$ \int_X |f_n| \le n^{p-1}\int_X |f_n|^p. $$ So arbitrarily close to $f\in L^p(X)$, in the $L^p$-norm, we can find an $f_n\in L^1\cap L^\infty$.

Note. This is true for ANY measure space $X$.