Under what conditions on $f$ does $\|f\|_r = \|f\|_s$ for $0 < r < s < \infty$. Question: 
If $f$ is a complex measurable function on $X$, such that $\mu(X) = 1$, and $\|f\|_{\infty} \neq 0$ when can we say that $\|f\|_r = \|f||_s$ given $0 < r < s \le \infty$?
What I know:
Via Jensen's inequality that,
$\|f\|_r \le \|f\|_s$ always holds true. For slightly more generality (for those interested, not because it's helpful here...) If $\mu(X) < \infty$ and $1 < r < s < \infty$ then, $\|f\|_r \le \|f\|_s \mu(X)^{\frac{1}{r} - \frac{1}s}$, follows from the Holder inequality.
Also, clearly $f \equiv 1$ is a solution.
What I've tried:
After not making any progress trying to find conditions for $\|f\|_s \le \|f \|_r$. I have tried to decompose $X$ to gain information. However, it leads to too many variables to be helpful, but maybe someone can improve my attempt, so here it is. Let $A = \{ x : |f(x)| < 1 \}, B = \{x : |f(x)| = 1 \}$, and $C = \{x : |f(x)| > 1\}$. Then, to find the necessary conditions we can set
\begin{align*} &\|f\|_s = \left( \int_{A} |f|^s d\mu + \mu(B) + \int_{C} |f|^s d\mu \right)^{1/s}\\ 
&= \left( \int_{A} |f|^r d\mu + \mu(B) + \int_{C} |f|^r d\mu \right)^{1/r} = \|f\|_r.
\end{align*}
However, I then proceeded to not make it anywhere that looked helpful later on.
Any new insight/suggestions are appreciated! Thanks.

Let $0 < r < s < \infty$. Using Holder's inequality,
$$\int_X |f|^r\ d\mu \le \left\{\int_X (|f|^{r})^{s/r} \ d\mu \right\}^{r/s}$$
i.e. $\|f\|_r \le \|f\|_s$. Equality holds iff $|f|$ is constant a.e.
If $s = \infty$, $$\|f\|_r^r = \|f\|_\infty^r$$
$$\int_X (\|f\|_\infty^r - |f|^r)\ d\mu = 0$$
so $|f| = \|f\|_\infty$ a.e., i.e. $|f|$ is constant almost everywhere.
