$\|f\|_p\to \|f\|_\infty$ under general assumptions I am trying to show that for nonnegative $f$ on $\mathbb{R}$ if $||f||_1<\infty$, we have
$$
\lim_{p\uparrow \infty}\|f\|_p=\|f\|_\infty
$$
I have tried to fiddle around with
$$
\left(\int_\mathbb{R}|f|^p|f|^{1-p} \right)^{1/p}<\infty
$$
but haven't found a nice way to get an estimate on $\int_E|f|^p$ from this. Honestly I am pretty uncomfortable with non finite measure spaces.
Any hints would be awesome! Any references for problems similar to this would also be awesome.
edit: this is not a duplicate of the linked question as 1) my measure space is not finite, 2) my function is in $L^1$ and not necessarily in $L^\infty$
 A: Let $(X,\mu)$ be a measure space, and let $f:X\to [0,\infty)$ be measurable.
Thm 1: If $\|f\|_\infty = \infty,$ then $\lim_{p\to \infty}\|f\|_p = \infty.$
Thm 2: If $\|f\|_\infty < \infty$ and $\|f\|_{p_0} < \infty$ for some $p_0 \in (0,\infty),$ then $\lim_{p\to \infty}\|f\|_p = \|f\|_\infty.$
Proof of Thm 1: Note that we have $\mu(\{f>C\}) >0$ for all $C>0$ because $\|f\|_\infty = \infty.$ Suppose $\mu(\{f>C\})=\infty$ for some $C>0.$ Then because
$$\tag 1 \int_X f^p\, d\mu \ge \int_{\{f>C\}} f^p\, d\mu \ge C^p\mu(\{f>C\})$$
for all $p\in (0,\infty),$ all integrals on the left of $(1)$ are $\infty.$ Thus all quantities of interest are stuck at the value $\infty,$ and we're done.
In the case $\mu(\{f>C\})<\infty$ for all $C>0,$ we see by $(1)$ that $\|f\|_p \ge C(\mu(\{f>C\}))^{1/p}.$ It then follows that $\liminf_{p\to \infty} \|f\|_p \ge C.$ Since $C$is arbitrarily large, $\liminf \|f\|_p =\infty,$ and we're done.
Proof of Thm 2: For any $p>p_0,$ we have
$$\int_X f^p\, d\mu = \int_X f^{p-p_0}\cdot f^{p_0}\, d\mu \le \|f\|_\infty^{p-p_0}\int_X f^{p_0}\, d\mu.$$
Now take $p$th roots to get
$$\|f\|_p \le (\|f\|_\infty)^{1-p_0/p}(\int_X f^{p_0}\, d\mu)^{1/p}.$$
This implies $ \limsup \|f\|_p \le \|f\|_\infty.$
On the other hand, if $0<\epsilon<\|f\|_\infty,$ we have $\mu(\{f>\|f\|_\infty-\epsilon\}) >0.$ Thus
$$\int_X f^p\, d\mu \ge \int_{\{f>\|f\|_\infty-\epsilon\}}f^p\, d\mu \ge (\|f\|_\infty-\epsilon\})^p\mu(\{f>\|f\|_\infty-\epsilon\}).$$
Take $p$th roots and let $p\to \infty$ to get $\liminf \|f\|_p \ge \|f\|_\infty-\epsilon.$ Since $\epsilon$ is arbitrarily small, we have $ \liminf \|f\|_p \ge \|f\|_\infty.$
We've shown $\|f\|_\infty \le \liminf \|f\|_p \le \limsup \|f\|_p \le \|f\|_\infty,$ so we're done.
A: Here is another proof to this problem:

Assume $0<\|f\|_\infty<\infty$ and $f\in L_r$ or some $r>0$. Then $|f|/\|f\|_\infty\leq1$ a.s.. For $p>r$
$$\frac{|f|^p}{\|f\|^p_\infty}\leq \frac{|f|^r}{\|f\|^r_\infty}\in L_1$$
hence $p\in E:=\{s: \|f\|_s<\infty\}$. Integrating on both side leads to
$$\frac{\|f\|_p}{\|f\|_\infty}\leq\Big(\frac{\|f\|_r}{\|f\|_\infty}\Big)^{r/p}\xrightarrow{p\rightarrow\infty}1$$
That is
$$\limsup_p\|f\|_p\leq \|f\|_\infty$$
By the Markov-Chebyshev inequality, for any $0<\alpha<\|f\|_\infty$
$$0<\alpha\big(\mu(|f|>\alpha)\big)^{1/p}\leq\|f\|_p$$
Hence $\alpha\leq\liminf_p\|f\|_p$ and so,
$\|f\|_\infty\leq\liminf_p\|f\|_p$.

If $\|f\|_p=\infty$ and $f\in L_r$ for some $r>0$ then $0<\mu(|f|<n)\leq\frac{1}{n^r}\|f\|_r<\infty$ and so
$$0<n\big(\mu(|f|>n)\big)^{1/p}\leq\|f\|_p\quad\text{for}\quad p\geq r$$
This implies $n\leq\liminf_p\|f\|p$ for any $n\in\mathbb{N}$.
A: Assume $\|f\|_\infty\lt\infty$.
For $\lambda\gt\|f\|_\infty$, $|\{x\in\Omega:|f(x)|\gt\lambda\}|=0$. Thus,
$$
\begin{align}
\|f\|_p^p
&=\int_\Omega|f(x)|^p\,\mathrm{d}x\\
&=p\int_0^\infty|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\lambda^{p-1}\,\mathrm{d}\lambda\\
&\le p\|f\|_\infty^{p-1}\int_0^\infty|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\mathrm{d}\lambda\\
&=p\|f\|_\infty^{p-1}\|f\|_1\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\limsup_{p\to\infty}\|f\|_p
&\le\lim_{p\to\infty}\left(p\|f\|_1\right)^{1/p}\|f\|_\infty^{(p-1)/p}\\
&=\|f\|_\infty\tag{2}
\end{align}
$$
If $\|f\|_\infty=\infty$, then $(2)$ is trivially true. Thus, $(2)$ is true whether $\|f\|_\infty\lt\infty$ or not. Therefore, we can remove the assumption that $\|f\|_\infty\lt\infty$.
For $\lambda\lt\|f\|_\infty$, $|\{x\in\Omega:|f(x)|\gt\lambda\}|\gt0$. Furthermore, for $\lambda\gt0$,
$$
\|f\|_p^p\ge|\{x\in\Omega:|f(x)|\gt\lambda\}|\,\lambda^p\tag{3}
$$
Therefore, for all $\lambda\lt\|f\|_\infty$,
$$
\begin{align}
\liminf_{p\to\infty}\|f\|_p
&\ge\lim_{p\to\infty}|\{x\in\Omega:|f(x)|\gt\lambda\}|^{1/p}\lambda\\
&=\lambda\tag{4}
\end{align}
$$
Since $(4)$ is true for all $\lambda\lt\|f\|_\infty$, we have
$$
\liminf_{p\to\infty}\|f\|_p\ge\|f\|_\infty\tag{5}
$$
Finally, $(2)$ and $(5)$ imply
$$
\lim_{p\to\infty}\|f\|_p=\|f\|_\infty\tag{6}
$$
