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Let $f$ be a function from $\{1,2,3,\dots,10\}$ to $\Bbb{R}$ such that $$\left(\sum\limits_{i=1}^{10}{\frac{|f(i)|}{2^i}}\right)^2=\left(\sum\limits_{i=1}^{10}{|f(i)|^2}\right)\left(\sum\limits_{i=1}^{10}{\frac{1}{4^i}}\right)$$ How many such $f$ are possible?

I used the Cauchy-Schwarz inequality to conclude that this condition would imply $2|f(1)|=2^2|f(2)|=\dots=2^{10}|f(10)|$, and hence there are uncountably many such functions possible.

However, I am not sure of this. Any help solving this question would be great.

Daniel Fischer
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algebraically_speaking
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  • Your reasoning looks good to me. And certainly, there are uncountably many such functions. But the answer is a bit more interesting if we state it as follows. "Up to a multiplicative constant, there is only one such function, namely $f(i) = \frac{1}{2^i}.$ If a function $g$ has the desired property, then we must have $g = \lambda f$ with an appropriate $\lambda \in \mathbb R.$" – jflipp Dec 15 '14 at 09:26
  • @jflipp You can chose the signs as well. – AlexR Dec 15 '14 at 09:33
  • @AlexR You're perfectly right. Unfortunately, I can't edit my old comment. So I just restate here that "the reasoning in the question is correct, the result can be formulated a bit more stringently by saying that up to a multiplicative constant there are exactly $2^{10}$ such functions". – jflipp Dec 15 '14 at 09:38

1 Answers1

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Assume that $f(i)>0$ (we'll see this doesn't change things much).

You have a space of $10$-dimensional vectors equipped with a standard dot product, with a condition that essentially says

$$(\vec{v}\cdot\vec{u})^2=\vec{v}^2\vec{u}^2$$ meaning that the vectors are parallel. So, there is a $1$-dimensional subspace of the original space, parallel to $\vec{u}=(1/2,1/4,1/8,\cdots)$ that solves the problem. Absolute values just mean the signs are arbitrary, so all possible solutions are

$$t(\pm 1/2,\pm 1/4,\pm 1/8,\cdots),\quad t\in \mathbb{R}$$

orion
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