Does there exist a constant $C>0$ such that for all simple functions $f$, $\int_1^\infty|f|\leq C(\int_1^\infty|f|^p)^{1/p}$?

Question

I'm trying to solve the following problem.

Let $p\in(0,\infty)$ be fixed. Determine, with justification, whether the following statement is true or false.

There exists a constant $C>0$ such that for all simple functions $f:[1,\infty)\to\mathbb{R}$, $$\int_1^\infty|f(x)|dx\leq C\left(\int_1^\infty|f(x)|^pdx\right)^{1/p}.$$

I think the statement is false. My thought is to proceed by supposing it is true. But I can't seem to arrive at a contradiction. Any hints regarding this are greatly appreciated. Thanks a lot.

Answer 1

Suppose such an inequality holds. Take $f=\chi_E$. We get $\mu (E)\leq C\mu(E)^{1/p}$ or $(\mu (E))^{1-\frac 1p} \leq C$. Is is easy tpo get a contradiciton by letting $\mu (E) \to 0$ or $\mu(E) \to \infty$.

Answer 2

Assume that it is true, for any $f\in L^{p}$, choose a sequence $(\varphi_{n})$ of simple functions such that $\varphi_{n}\rightarrow f$ in $L^{p}$ and $\varphi_{n}(x)\rightarrow f(x)$ a.e.

Fatou's lemma implies that \begin{align*} \int|f(x)|\leq\liminf\int|\varphi_{n}(x)|dx\leq C\liminf\left(\int|\varphi_{n}(x)|^{p}\right)^{1/p}\leq C\left(\int|f(x)|^{p}\right)^{1/p}. \end{align*}

For $p>1$, choose $f(x)=\dfrac{1}{x^{\eta}}$, where $0<\eta<1$ and $\eta p>1$, then \begin{align*} \int_{1}^{\infty}|f(x)|^{p}dx<\infty \end{align*} but \begin{align*} \int_{1}^{\infty}|f(x)|dx=\infty. \end{align*}