How to bound $m(A)^{-1}\int_A|f(x) - c|^tdx$ above by $\left(m(A)^{-1}\int_A|f(x) - c|dx\right)^t$ for $0 < t < 1$ and $A$ a compact set

Question

Let $m$ denote the Lebesgue measure on $\mathbb{R}^n$, $f$ be a locally integrable function in $\mathbb{R}^n$, let $0 < t < 1$, $A\subset\mathbb{R}^n$ be a compact subset and $c$ some finite constant. I am trying to bound $m(A)^{-1}\int_A|f(x) - c|^tdx$ above by $\left(m(A)^{-1}\int_A|f(x) - c|dx\right)^t$, but so far have been unsuccessful. I initially wanted to apply Jensen's inequality, which would immediately give the result, but the only source I know for the proof of Jensen's ineq. uses the Hölder's inequality and requires $1 \leq t$. What should I do?

Answer 1

Note that $\int_A |f(x)-c|^t dx = \int_A |f(x)-c|^t 1 dx$, so applying Holder's inequality with $p = \frac 1t > 1$ and $q = \frac{1}{1-t}$ we have \begin{align*} \int_A |f(x)-c|^t 1 dx &\le \left( \int_A |f(x)-c|dx \right)^t \left(\int_A 1 dx\right)^{1-t} \\ &= \left( \int_A |f(x)-c|dx \right)^t m(A)^{1-t} \\ &= \left( m(A)^{-1}\int_A |f(x)-c|dx \right)^t m(A). \end{align*} Thus \begin{align*} m(A)^{-1}\int_A |f(x)-c|^t 1 dx &\le m(A)^{-1}\left( m(A)^{-1}\int_A |f(x)-c|dx \right)^t m(A) \\ &= \left( m(A)^{-1}\int_A |f(x)-c|dx \right)^t. \end{align*}

Answer 2

Since $1/t\geq 1$, Jensen’s inequality gives

$$\left(\frac1{m(A)}\int_A |f(x)-c|^t dx\right)^{1/t} \leq \frac1{m(A)} \int_A (|f(x)-c|^t)^{1/t} dx$$

Which is exactly the desired result. Essentially, since $x\mapsto x^t$ is concave for $0<t<1$, the inequality from Jensen is reversed.