Let $f \in C^1([0;1],\mathbb{R})$ such that $f(1)=0$.

$$\text{Prove that} \qquad \int_{0}^{1}|f(x)|dx \leq \int_{0}^{1}x|f'(x)|dx$$

My attempt:

\begin{align} \int_{0}^{1}x|f'(x)|dx = \int_{0}^{1}|xf'(x)|dx &\geq \left | \int_{0}^{1}xf'(x)dx \right | \\ &= \left | xf(x)|^1_0 - \int_{0}^{1}f(x)dx \right | = \left | \int_{0}^{1}f(x)dx \right | \end{align}

I'm stuck here because $\left | \int_{0}^{1}f(x)dx \right | \leq \int_{0}^{1}|f(x)|dx$ while I want something $\geq \int_{0}^{1}|f(x)|dx$

  • 2
    $\begingroup$ It is my personal opinion that people should try harder before they ask others to solve their homework problems for them. $\endgroup$ – uniquesolution Apr 13 at 18:44

\begin{align*} \int _0 ^1 |f(x)| dx &= \int _0 ^1 \left| - \int _x ^1 f'(t) dt\right| dx \leq \int _0 ^1 \int _x ^1 |f'(t)| dt dx \\ &= \int _0 ^1 \int _ 0 ^t |f'(t)| dx dt = \int_0 ^1 t|f'(t)| dt = \int_0 ^1 x|f'(x)| dx \end{align*}

  • 2
    $\begingroup$ Nice solution, if Fubini's Theorem is available to the OP. $\endgroup$ – Ted Shifrin Apr 13 at 19:05
  • $\begingroup$ Very nice solution.(+1) $\endgroup$ – DINEDINE Apr 13 at 20:39

I used a double integral also, but without the need to switch the limits of integration:

set $F(x)=\int^x_0|f'(t)|dt.$

Then, an integration by parts and a calculation proves the claim :

$\int^1_0x|f'(x)|dx=1\cdot F(1)-0\cdot F(0)-\int^1_0F(x)dx=$



$\int^1_0\int^1_x|f'(t)|dtdx\ge \int^1_0\left|\int^1_xf'(t)dt\right|dx=\int^1_0|f(x)|dx$

$^*$ I would like to see a proof that does not use double integrals


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