# Proof no nonnegative function satisfies 3 integrals (3 $L^2$ norms derived from the inner product on $C[0,1]$)

$$V$$ is the real inner product space $$C_\mathbb{R}[0,1]$$ and $$a \in [0,1]$$. Prove there is no nonnegative function $$f \in V$$ such that $$\int_0^1 f(x) \, dx = 1,$$ $$\int_0^1 xf(x) \, dx = a,$$ $$\mathrm{and} \int_0^1 x^2 f(x) \, dx = a^2$$

We have that the norm derived from the $$L^2$$ inner product on $$C[0,1]$$ is the $$L^2$$ norm $$\langle g, g \rangle = \Vert g \Vert^2 = \int _0^1 |g(t)|^2 \, dt$$ We also have that the Cauchy-Schwartz inequality states that: $$\langle u,v \rangle \leq \Vert u \Vert \Vert v \Vert$$ (With equality if and only if $$u$$ and $$v$$ are scalar multiples.) Here, for an inner product of something with itself, C.-S. becomes: $$\langle g,g \rangle = \Vert g \Vert^2$$ So given the above 3 integrals, we have $$g_1 = f(x)$$ $$g_2 = xf(x)$$ $$g_3 = x^2f(x)$$ And accordingly, $$\Vert g_1 \Vert^2 = 1, \, g_1 = \sqrt{f(x)}$$ $$\Vert g_2 \Vert^2 = a, \, g_2 = \sqrt{xf(x)}$$ $$\Vert g_3 \Vert^2 = a^2, \, g_3 = \sqrt{x^2f(x)}$$

However, after several attempts, I can't seem to finish the proof along this line of thought. Perhaps I'm squaring or square-rooting something incorrectly? I suspect that when we square $$a$$ from the second integral and try to equate things with $$a^2$$ from the third, something goes wrong with Cauchy-Schwartz. But how to bring it home?

• Given the information, $f$ can be seen as a probability density function of a r.v $X\in [0,1]$ such that $\Bbb E[X]=a, \Bbb E[X^2]=a^2$. Then, $\text{var}(X)=\Bbb E[X^2]-\Bbb E[X]^2=0$, which implies $\Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf. – Song Mar 22 at 4:09

The three integrals imply that $$\int_0^1(x-a)^2 f(x) dx=0$$, which implies that $$f(x)\equiv 0$$. But this does not satisfy the first integral.