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let $f:[0,1]\longrightarrow R$ be a differentiable function with continuous derivative such that $f(1)=0$,show that:

$$4\int_{0}^{1}x^2|f'(x)|^2dx\ge\int_{0}^{1}|f(x)|^2+\left(\int_{0}^{1}|f(x)|dx\right)^2$$

I think it can be use Cauchy-schwarz inequality

$$\int_{a}^{b}f^2(x)dx\int_{a}^{b}g^2(x)dx\ge\int_{a}^{b}(f(x)g(x))^2dx$$

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1 Answer 1

up vote 5 down vote accepted

Let

$$c := \int_0^1 f(x) \, dx \tag{1}$$

Without loss of generality, we may assume

$$\int_0^1 (f(x)+c)^2 \, dx >0$$

otherwise $f=-c$, hence $f=0$ (since $f(1)=0$, by assumption) and in this case the inequality is trivially satisfied.

Integration by parts yields

$$\begin{align*} \int_0^1 (f(x)+c)^2 \, dx &= \bigg[x \cdot (f(x)+c)^2\bigg]_0^1 - 2 \int_0^1 x \cdot (f(x)+c) \cdot f'(x) \\ &= c^2 + 2 \int_0^1 -(f(x)+c) \cdot f'(x) \cdot x \end{align*}$$

where we used that $f(1)=0$. By applying Jensen's inequality, we obtain

$$\int_0^1 |f(x)+c|^2 \, dx -c^2 \leq 2 \sqrt{ \int_0^1 |f(x)+c|^2 \, dx} \cdot \sqrt{\int_0^1 x^2 \cdot f'(x)^2 \, dx}$$

i.e.

$$\sqrt{\int_0^1 |f(x)+c|^2 \, dx} - \frac{c^2}{\sqrt{\int_0^1 |f(x)+c|^2 \, dx}} \leq 2 \sqrt{\int_0^1 x^2 \cdot f'(x)^2 \, dx} $$

Squaring both sides yields

$$\int_0^1 |f(x)+c|^2 \, dx - 2c^2 + \frac{c^4}{\int_0^1 |f(x)+c|^2 \, dx} \leq 4 \int_0^1 x^2 \cdot f'(x)^2 \, dx \tag{2}$$

Note that by definition

$$\int_0^1 |f(x)+c|^2 \, dx = \int_0^1 f(x)^2 \, dx +2c \underbrace{\int_0^1 f(x) \, dx}_{c} + c^2 \stackrel{(1)}{=} \int_0^1 f(x)^2 \, dx + 3 \left( \int_0^1 f(x) \, dx \right)^2 $$

Thus, $(2)$ is equivalent to

$$\int_0^1 |f(x)|^2 \, dx + \left( \int_0^1 f(x) \, dx \right)^2 + \underbrace{\frac{c^4}{\int_0^1 |f(x)+c|^2 \, dx}}_{\geq 0} \leq 4 \int_0^1 x^2 \cdot f'(x)^2 \, dx$$

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