If $g:[a,b] \to \mathbb{R}^m$ is continuous and $x_0 \in [a,b]$, prove $|\int_a^b g(x)dx| \geq \int^a_b|g(x)|dx-2\int^a_b|g(x)-g(x_0)|dx$.

Question

This is an exercise from Leoni’s book on Sobolev spaces. It seems elementary but the proof is eluding me despite my repeated attempts to use the reverse triangle inequality and the mean value theorem. There might be a trick I haven’t thought of yet.

Answer 1

The idea, as suggested in the comments, is to use the inequality: $$ \int_a^b \left\lvert g(x) - \frac1{b-a} \int_a^b g(y) \,\mathrm{d}y\right\rvert \,\mathrm{d}x \leq 2 \int_a^b \lvert g(x) - c \rvert \,\mathrm{d}x \tag{$\dagger$}$$ for all $c \in \mathbb R^m$. If this holds, we can then apply this with $c = g(x_0)$ to estimate \begin{align} \int_a^b \lvert g(x) \rvert\,\mathrm{d}x &\leq \int_a^b \left\lvert g(x) - \frac1{b-a} \int_a^b g(y) \,\mathrm{d}y\right\rvert \,\mathrm{d}x + \int_a^b\left\lvert\frac1{b-a} \int_a^b g(y) \,\mathrm{d}y\right\rvert\,\mathrm{d}x \\ &\leq 2 \int_a^b \lvert g(x) - g(x_0)\rvert \,\mathrm{d}x + \left\lvert \int_a^b g(y) \,\mathrm{d}y\right\rvert, \end{align} which we can rearrange to arrive at the desired estimate.

To show $(\dagger)$, as you say in the comments you estimate the left-hand side by a double integral, and to bound this one uses the triange inequality to split $$ \lvert g(x) - g(y) \rvert \leq \lvert g(x) - c \rvert + \lvert c - g(y) \rvert. $$ From here one can integrate out the free variable in the respective terms to conclude. In detail one has \begin{align} \int_a^b \left\lvert g(x) - \frac1{b-a} \int_a^b g(y) \,\mathrm{d}y\right\rvert \,\mathrm{d}x &\leq \frac1{b-a} \int_a^b \int_a^b \lvert g(x) - g(y) \rvert \,\mathrm{d}y \,\mathrm{d}x\\ &\leq \frac1{b-a}\int_a^b\int_a^b \lvert g(x) - c \rvert + \lvert c - g(y) \rvert \,\mathrm{d}y \,\mathrm{d}x \\ &= \left(\frac1{b-a}\int_a^b\,\mathrm{d} y\right)\int_a^b \lvert g(x) - c \rvert \,\mathrm{d}x + \left(\frac1{b-a}\int_a^b\,\mathrm{d} x\right)\int_a^b \lvert c - g(y) \rvert \,\mathrm{d}y \\ &= 2 \int_a^b \lvert g(x) - c \rvert \,\mathrm{d}x, \end{align} as required. Here note that the two integrals in the penultimate line are the same, just that the variable of integration changes from $x$ to $y$.