Let f:[0,1]→R be a differentiable function, with integrable derivative on [0,1], such that f(1)=0. Prove that ∫01(xf′(x))2dx≥12⋅(∫01xf(x)dx)2.
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Let g:[0,1]→R defined by g(x)=xf(x), for x∈[0,1]. We have g(0)=0=g(1), g′(x)=f(x)+xf′(x), so that: ∫01x2(f′(x))2dx=∫01(g′(x)−f(x))2dx=∫01(g′(x))2dx−2∫01g′(x)f(x)dx+∫01(f(x))2dx=∫01f2(x)dx+∫01(g′(x))2dx−−2(f(1)g(1)−f(0)g(0))+2∫01g(x)f′(x)dx, implying ∫01x2(f′(x))2dx==∫01f2(x)dx+∫01(g′(x))2dx+∫01x⋅(2f(x)f′(x))dx=∫01f2(x)dx+∫01(g′(x))2dx+∫01x⋅(f2(x))′dx=∫01f2(x)dx+∫01(g′(x))2dx+(1⋅f2(1)−0⋅f2(0))−∫01f2(x)dx=∫01f2(x)dx+∫01(g′(x))2dx−∫01f2(x)dx=∫01(g′(x))2dx.
(∫01(2x−1)⋅g′(x)dx)2≤∫01(2x−1)2dx⋅∫01(g′(x))2dx==61⋅((2⋅1−1)3−(2⋅0−1)3)⋅∫01x2(f′(x))2dx=31⋅∫01x2(f′(x))2dx. This gives ∫01(xf′(x))2dx≥3⋅(∫01(2x−1)⋅g′(x)dx)2==3⋅((2⋅1−1)g(1)−(2⋅0−1)g(0)−2∫01g(x)dx)2==12(∫01xf(x)dx)2, concluding the proof.