Show that $|a|\leq \int_{-1}^{1}\,|ax+b|\,\text{d}x$

Question

Show that $\displaystyle|a|\leq \int_{-1}^{1}\,|ax+b|\,\text{d}x$.

I did this problem but in a way that I do not like. I just divided it into many cases. Assumed $a>0$ and then did cases depending on where $-b/a$ is. However, this seems very computational and there must be a simpler way, at least i hope. This problem arose when considering the linear functional from $Span\{1,x\}$ with $L^1$ norm to $f'(0)$. I wanted to show that this operator is bounded which leads to the following inequality.

Answer 1

\begin{align} \int_{-1}^1|ax + b| {\rm d}x&=\int_{-1}^1\big|(ax + b)\,\text{sign}(x)\big| {\rm d}x \\&\ge \left|\int_{-1}^1 (ax + b)\,\text{sign}(x) {\rm d}x\right|\\ &= |a|. \end{align}

(I think this is the first time I use the inequality $\displaystyle\int_a^b|f| \ge \left|\int_a^bf\right|$ in this direction!)

Edit by Batominovski. Please undelete this. Your idea works. It just needs a small twist. I fixed that for you. (You can delete this remark. I am ok with that.)

Edit by Aryaman. The above solution is mainly thanks to Batominovski, who fixed a rather silly calculation of mine. Thank you!

Answer 2

The case $a=0$ is obvious so one may assume that $a\neq 0$. Dividing both sides by $|a|$, it suffices to prove that $$1\leq \int_{-1}^1|x+c|~dx,$$ for any real $c$. By geometry (using area) it is easy to see that $$|c|>1\Rightarrow \int_{-1}^1|x+c|~dx>2,$$ while for $|c|\leq 1,$ one has $$\int_{-1}^1|x+c|~dx=\frac 12((1+c)^2+(1-c)^2)=1+c^2\geq 1.$$ QED

Answer 3

From a graphical point of view, in the case $b=0$, the integral is the green area in the figure, and is obviously equal to $a$ (supposing $a>0$).

When $b>0$ we have the following figure, and the area changes

If we compare the two figures, we see that moving from the first to the second case the area increase of the green part and decrease of the red part, but the green part is bigger, so that the area of the second case is bigger with respect to the first case.

In the case $b<0$ the graphic move to the right, instead of to the left, but we can make the same considerations and see that the area increase with respect to the case $b=0$, so in any case the area is greater or equal to $a$.