# Integral inequality with specific condition

$$f(x)$$ is integrable on $$[0;1]$$ and $$\int\limits_0^1 f(x) dx = \int\limits_0^1 xf(x) dx = 1$$. Prove that $$\int\limits_0^1(f(x))^2 dx\geq4$$ I tried to use Cauchy–Schwarz integral inequality. I tried a lot of transformations, but none of them are successful.

• Hint: Suppose $f = \alpha_1 + \alpha_2 x$. The conditions say that $\langle f,1 \rangle = \langle f, x \rangle = 1$. The quantity you want to bound is $\|f\|^2 = \langle f, f \rangle$. Some algebra here should prove the claim for these $f$, then argue that any additional terms in $f$ only add to the norm. Apr 12 at 17:59
• @whpowell96 I didn't get how exactly scalar product $\langle a, b \rangle = \int\limits_0^1 ab dx$ might help me. I tried to use axioms, but it looks like it doesn't work here. Apr 12 at 18:53
• I have elaborated my comment into a proof Apr 12 at 19:07
• Something else : please validate the (good) answers to the questions you ask. For example, you haven't done it for the question you asked almost two months ago about sequences. You have to do it : his is how this site works. Apr 12 at 20:57

## 2 Answers

Note $$\begin{eqnarray} &&\int_0^1(f(x)+2-6x))^2 dx\\ &=&\int_0^1(f^2(x)+36x^2+4+4f(x)-12xf(x)-24x)dx\\ &=&\int_0^1f^2(x)dx-4\geq0 \end{eqnarray}$$ and hence $$\int_0^1f^2(x)dx\geq4$$ and "$$=$$" holds if and only if $$f(x)=-2+6x$$.

• Very nice solution Apr 12 at 22:02
• Thank you very much! Apr 12 at 22:26

Let $$V$$ be the subspace of $$L^2[0,1]$$ spanned by $$\{1,x\}$$. We are given that $$\langle f,1 \rangle = \langle f,x \rangle = 1$$. For simplicity, we will orthogonalize this basis by defining $$e_1 = 1$$, $$e_2 = \sqrt{12}(x - 1/2)$$. Indeed, we have $$\|e_1\| = \|e_2\| = 1$$, $$\langle e_1, e_2 \rangle=0$$, as well as $$\langle f, e_1 \rangle = 1$$ and $$\langle f, e_2 \rangle = \sqrt{12} \langle f, x \rangle - \frac{\sqrt{12}}{2}\langle f, 1 \rangle = \frac{\sqrt{12}}{2} = \sqrt{3}$$. Therefore, $$f = e_1 + \sqrt{3} e_2 + g$$, where $$g \in V^\perp$$. Since $$\{e_1,e_2\}$$ is an orthonormal basis of $$V$$ and $$g\in V^\perp$$, we have $$\|f\|^2 = 1^2 + (\sqrt{3})^2 + \|g\|^2 = 4 + \|g\|^2 \geq 4,$$ where equality is obtained iff $$f\in V$$.