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.

  • 4
    $\begingroup$ In Hölder's inequality, you pair $\lvert f\rvert$ up with $1$. Looking at when you have equality in Hölder's inequality, the answer is $\lVert f\rVert_r = \lVert f\rVert_s$ for $0 < r < s \leqslant \infty$ if and only if $\lvert f\rvert$ is constant [a.e.]. $\endgroup$ Feb 14, 2014 at 21:31

2 Answers 2


A look at the proof of Jensen's inequality is all you really need; there is no need for (more sophisticated) Hölder's inequality.

For simplicity, scale $f$ so that $\|f\|_r=1$. Let $g=|f|^r$. Jensen's inequality says $\int_X g^{p}\ge 1$ (for $p=s/r>1$), which is just the result of integrating the pointwise inequality $$g^{p} \ge 1+p(g-1) \tag{1}$$ over $X$. Inequality (1) expresses the convexity of the function $g\mapsto g^p$: its graph lies above the tangent line at $g=1$. Recall that an integral of a nonnegative function is zero only when the function is zero a.e. Thus, if we have
$$\int_X g^p = \int_X (1+p(g-1)) \tag{2}$$ then (1) holds as equality a.e. But (1) holds as equality only when $g=1$; this is the only point at which the tangent line to $g\mapsto g^p$ meets the graph of the function. Therefore, $g=1$ a.e.

In terms of $f$, which means $|f|$ being constant a.e., as Daniel Fischer noted.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.