# How prove this integral inequality $\min_{x\in[0,1]}f(x)\ge-\int_{0}^{1}|f'(x)|dx$

let $f(x)$ can change sign in $x\in [0,1]$ and is continuous derivative function, show that

$$\min_{x\in[0,1]}f(x)\ge-\int_{0}^{1}|f'(x)|dx$$

My try:since $f(x)$ change sign in $x\in [0,1]$,

then there exsit $\xi\in [0,1]$ such $f(\xi)=0$. then

$$\Longleftrightarrow \min_{x\in [0,1]}f(x)-f(\xi)\ge \int_{0}^{1}|f'(x)|dx$$ then let $$f(a)=\min_{x\in[0,1]}f(x)$$ so $$\Longrightarrow f(\xi)-f(a)\le\int_{0}^{1}|f'(x)|dx$$

then How to prove this inequality?

Thank you(maybe this problem have some methods) for your nice methods.

If $x\in[0,1]$ then \begin{align} f(x)&=\int_0^xf'(t)\,dt\\ &\ge-\Bigl|\int_0^xf'(t)\,dt\Bigr|\\ &\ge-\int_0^x|f'(t)|\,dt\\ &\ge-\int_0^1|f'(t)|\,dt. \end{align}

## Edit

I just realized that I made a mistake, since $f(x)=f(0)+\int_0^xf'(t)\,dt$. The above is valid only if $f(0)=0$. If $f$ does not change sign, then the inequality may fail, as the example $f(x)=-1$ shows.

If $f(\xi)=0$ for some $\xi\in[0,1]$ and $\xi\le x\le1$ then \begin{align} f(x)&=\int_\xi^xf'(t)\,dt\\ &\ge-\Bigl|\int_\xi^xf'(t)\,dt\Bigr|\\ &\ge-\int_\xi^x|f'(t)|\,dt\\ &\ge-\int_0^1|f'(t)|\,dt. \end{align} Similarly if $0\le x\le\xi$.

• Hello, this is $f(0)\neq 0$ – math110 Nov 12 '13 at 17:26
• I ha edited my answer. – Julián Aguirre Nov 12 '13 at 17:32

Let $a\in[0,1]$ such that $f(a)=\min_{x\in[0,1]}f(x)$. since $f$ change the sign there is $z\in[0,1]$ such that $f(z)= 0$

We have $$0= f(z) \ge f(a)=\min_{x\in[0,1]}f(x)$$ hence,

$$0\le -f(a) = f(z) -f(a) =\int_a^zf'(t)dt \le \left|\int_a^zf'(t)dt\right| \le \int_0^1\left|f'(t)\right|dt$$ that is $$f(a)=\min_{x\in[0,1]}f(x) \ge - \int_0^1\left|f'(t)\right|dt$$