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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.

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    $\begingroup$ 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.]. $\endgroup$ Commented Feb 14, 2014 at 21:31

2 Answers 2

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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.

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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.

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