About the proof of the inequality: $\int_0^1 |f(x)|dx\le\text{max}\left(\int_0^1 |f’(x)|dx, \left\vert\int_0^1 f(x)dx\right\vert\right)$

Question

设f(x)导数处处连续，证明： ∫|f(x)|dx≤MAX{∫|f'(x)|dx,|∫f(x)dx|}. 三个都为定积分 x from 0 to 1.

我用到了MAX{A,B}=(A+B+|A-B|)/2.加上|A-B|≥|A|-|B|. 但是和要证的不等号方向相反。

Translation:

Let $f$ be a function such that $f'$ is continuous everywhere, prove

$$\int_0^1 |f(x)|dx\le\text{max}\left(\int_0^1 |f’(x)|dx, \left\vert\int_0^1 f(x)dx\right\vert\right)$$

I have used $\text{max}(A,B)=\frac{A+B+|A-B|}2$ and $|A-B|\ge|A|-|B|$, however what I need to prove has the reversed inequality sign.

Answer 1

If $f>0$ or $f<0$ everywhere then the inequality is obviuous. Otherwise $f(c)=0$ for some $c$. Let $g =\frac f {R.H.S}$. Note that $\int |g'(t)|\, dt \leq 1$ and $|\int g(t)\, dt | \leq 1$. Now $|g(x)-g(c)|=|\int_c^{x} g'(t)\, dt| \leq \int_{0}^{1} |g'(t)|\, dt \leq1$ so $|g(x)| \leq 1$ for all $x$ and $\int_0^{1} |g(x)|\, dx \leq 1$. If you write $g$ in terms of $f$ you get the desired inequality.