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In a linear vector space that is the Euclidean space $\mathbb{E}_{\infty}$, we have the Cauchy-Schwarz inequality

$$ |\langle x,y \rangle| \leq |x| |y|,$$

where both $x,y \in\mathbb{E}_{\infty}$. Explicitly $x=(\xi_{1},\xi_{2},\cdots)$ and $y=(\eta_{1},\eta_{2},\cdots)$, and we assume here that both are finite in length (i.e. the series for $|x|^{2}$ and $|y|^{2}$ converge).

I am familiar with several proofs of this inequality. However, I am unable to follow the particular approach that Friedman indicates in one of the problems (Problem 1.2, p.6) in his book (also I noticed it's posted online here) where he says that we can prove the inequality by using the result

$$ |\alpha x + \beta y|^{2}=\langle \alpha x + \beta y, \alpha x + \beta y\rangle = \alpha^{2} \langle x,x \rangle +2 \alpha \beta \langle x,y \rangle + \beta^{2} \langle y,y \rangle,$$

which holds for any $\alpha,\beta$ scalars, and by putting

$$ x_{n}=(\xi_{1},\xi_{2},\cdots,\xi_{n},0,0,\cdots), $$ $$ \alpha = |y|^{2}, $$ $$ \beta=\langle x_{n},y\rangle, $$

to prove that $\langle x_{n},y\rangle \leq |x_{n}| |y|$.

How do we proceed using this specific approach (not other approaches) to explicitly reach the inequality?

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1 Answer 1

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I expect it's a typo, and $\ \beta\ $ should be $\ -\big\langle x_n, y\big\rangle\ $. You'll then get (with $\ \alpha=|y|^2\ $) \begin{align} \big|\alpha x_n+\beta y|^2&=|y|^4|x_n|^2-2|y|^2\big\langle x_n, y\big\rangle^2+ |y|^2\big\langle x_n, y\big\rangle^2\\ &=|y|^4|x_n|^2-|y|^2\big\langle x_n, y\big\rangle^2\ , \end{align} or \begin{align} \big\langle x_n, y\big\rangle^2&= |y|^2|x_n|^2-\frac{\big|\alpha x_n+\beta y|^2}{|y|^2}\\ &\le |y|^2|x_n|^2\ , \end{align} from which the Cauchy-Schwarz inequality follows immediately.

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  • $\begingroup$ Ah, that explains it, I think! Thanks! But why did he bother using $x_{n}$ at all here (instead of $x$)? Couldn't we just do the same with $x$ (since it is assumed to have finite length, so the series for $|x|^{2}$ converges)? $\endgroup$
    – user135626
    Commented Nov 9, 2020 at 19:56
  • $\begingroup$ I presume it's because if you use $\ x\ $ instead, you have to show that the series for $\ \langle x,y\rangle\ $ converges to justify your expansion of $\ |\alpha x+\beta y|^2\ $. While this isn't hard, it still has to be done for the proof to be watertight. This also applies to Friedman's proof, however, and the convergence doesn't follow immediately from his final inequality, so I guess my assertion that the Cauchy-Schwarz inequality follows "immediately" from it is a mild exaggeration. $\endgroup$ Commented Nov 10, 2020 at 0:02
  • $\begingroup$ But would you agree that if all series are assumed to converge, then it wouldn't make a difference using $x$ or $x_{n}$? $\endgroup$
    – user135626
    Commented Nov 10, 2020 at 0:25
  • $\begingroup$ Yes, if you make that assumption. While Friedman's exposition might be a little confusing, his purpose seems to be to get you to prove that $\ \mathbb{E}_\infty\ $ is a linear space via the exercises $\mathbf{1.1}$ to $\mathbf{1.3}$. He apparently expects that you won't prove that all the necessary series converge until you get to exercise $\mathbf{1.3}$. $\endgroup$ Commented Nov 10, 2020 at 1:04
  • $\begingroup$ It begs the question why didn't he treat $y$ similarly (i.e. why not used $y_{n}$)? Sure, when one multiplies with $x_{n}$ it will be similarly truncated, but in other terms $y$ is on its own, yet he only took $x_{n}$ not $y_{n}$. I wonder whether he is alluding to something else. $\endgroup$
    – user135626
    Commented Nov 10, 2020 at 1:10

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