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$V$ 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?

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    $\begingroup$ 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. $\endgroup$ – Song Mar 22 at 4:09
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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.

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