Lebesgue space inclusion on half real line

Question

Assume that $f \in L^1 ([0,\infty))$ is a real continuous function on $[0,\infty)$.

Can we then conclude that $f \in L^p ([0,\infty))$ for $p \in [1, \infty)$?

Answer 1

This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then, $$ \int g_n(x)\ dx = \frac12 n \cdot 2n^{-3} = n^{-2}. $$ Since for $n \geq 2$ functions $g_n$ have disjoint support, if we define $f = \sum_{n \geq 2} g_n$ then $f \in L^1([0, \infty))$. However, if we take $p = 2$, we have that $$ \int g_n(x)^2\ dx \geq \int_{n + \frac12 n^{-3}}^{n + \frac32 n^{-3}} (\frac12 n)^2 \ dx = \frac14 n^{-1}. $$ This implies $\int f^2 \geq \frac14 \sum_{n \geq 2} n^{-1} = \infty$.