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I would appreciate if somebody could help me with the following problem

Q: prove that

$$\int_{0}^{1}|f(x)|dx \leq \int_{0}^{1}|f'(x)|dx$$ where $f'(x)$ is continuous on $(0,1)$, and $f(0)=0$.

Young
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4 Answers4

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By the fundamental theorem of Calculus, $$ \left|f(x)\right| = \left|f(x) - f(0)\right| = \left|\int_0^x f'(t) dt\right| \le \int_0^x \left|f'(t)\right|dt \le \int_0^1 \left|f'(t)\right|dt. $$

The result now follows by integrating with respect to $x$ from $0$ to $1$.

Ayman Hourieh
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Using the fundamental theorem of calculus and $f(0)=0$ we obtain $f(x)=\int_0^x f'(t)dt$ and hence $$\int_0^1 |f(x)|dx=\int_0^1 \left|\int_0^x f'(t)dt\right|dx \leq \int_0^1 \int_0^x |f'(t)|dt\ dx\leq \int_0^1 \int_0^1 |f'(t)|dt\ dx = \int_0^1 |f'(t)|dt$$ since $|f'(t)|\geq 0$.

dinosaur
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Partial answer: If you suppose $f'\geq 0$, then $f$ is non decreasing, so $f \geq f(0) = 0$. Hence $$\int_0^1 f(x) dx \leq \int_0^1 \max f dx = f(1) = f(1) - f(0) = \int_0^1 f'(x) dx $$

If you suppose $f' \leq 0$ you get the same result.

Now maybe one can work separately on the intervals where $f'$ is $\geq 0$ or $\leq 0$ alternately, using the fact that $f'$ is continuous.

Crostul
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For any $f$, let us split the interval $[0,1]$ into subintervals such that $f'(x)$ has a constant sign. We can define a function $g(x)$ by "mirroring" all decreasing sections, i.e. $$g(x)=\int_0^x|f'(x)|dx.$$ We have $\int_0^x f'(x)dx\le\int_0^x|f'(x)|dx$, so that $f(x)\le g(x)$, and we have $|f'(x)|=g'(x)$. It is enough to consider $g(x)$.

By the mean value theorem, $$\int_0^1g(x)dx\le g(1),$$ and $$\int_0^1g'(x)dx=g(1).$$

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  • @ Yves Daoust I think that you cannot in general split the interval [0,1] into subintervals such that f′(x) has a constant sign on these subintervals. – Kelenner Aug 03 '14 at 14:27
  • @Kelenner: can you elaborate ? Actually, this is implicit in the definition of $g$. –  Aug 03 '14 at 14:32
  • @ Yves Daoust :You say $f$, not $g$. Take $\displaystyle h(x)=(x-\frac{1}{2})^4\sin(\frac{1}{(x-1/2)})$ for $x\not =1/2$, $h(1/2)=0$; $h$ is clearly continuous on $[0,1]$; put $H(x)=\int_0^x h(t)dt$, we have $H^{\prime}(x)=h(x)$, and for any interval (non reduced to a point) containing $1/2$, $h$ takes strictly positive and strictly negative values. – Kelenner Aug 03 '14 at 14:36
  • @Kelenner: this is just a loose statement to give $g$ an intuitive explanation. This does not impair the definition of $g$ and the validity of the corresponding inequalities. –  Aug 03 '14 at 14:43