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Let $\{x_n\}_{n∈N},\ \{y_n\}_{n∈N}$ be two orthonormal sequences in an Hilbert space $H$. Assume that $\lim_{n→+∞} \langle x_n, y_n \rangle_H = 1$. Prove that $\lim_{n→+∞} ||x_n − y_n||_H = 0.$

From the text I suppose the sequences are in $\ell^2$, so:

$\begin{align*} ||x_n − y_n||_2 = \langle x_n − y_n,x_n − y_n \rangle &= \langle x_n,x_n \rangle - \langle x_n,y_n \rangle - \langle y_n,x_n \rangle + \langle y_n,y_n \rangle\\ &= \langle x_n,x_n \rangle - 2\langle x_n,y_n \rangle + \langle y_n,y_n \rangle \end{align*}$

$\big($in case the sequences are made of real numbers, does all these inner products be actually the usual products? i.e. $\langle x_n,y_n \rangle = x_ny_n?)$

Since the sequences are in $\ell^2$ they converge to $0$, i.e. $\lim_{n\to+\infty} x_n = 0$ and same for $y_n$, so I would say that:

$\lim_{n\to+\infty}\langle x_n,x_n \rangle = 0,\quad \lim_{n\to+\infty}\langle y_n,y_n \rangle = 0$.

Then I obtain $\lim_{n\to+\infty} ||x_n − y_n||_2 = -2 \lim_{n\to+\infty} \langle x_n,y_n \rangle = -2\ (\ne 0),$ since by hypothesis $\lim_{n→+∞} \langle x_n, y_n \rangle = 1.$

Moreover, I'm trying to figure out how can be that $\lim_{n→+∞} \langle x_n, y_n \rangle = 1$. I don't know if in this case we can apply the Bessel inequality because we have two orthonormal sequences, in case we can I would say : $\sum_{n=1}^{\infty} |\langle x_n,y_n \rangle|^2 \le ||x||_2^2=1$ since the sequence is orthonormal. Does not follow from this that the series converges and so its term tends to zero?

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    $\begingroup$ By Cauchy Schwartz $|\langle x_n,y_n\rangle | \leq \|x_n\|\|y_n\|$ so if $\|x_n\|\to 0$ and $\|y_n\|\to 0$, then $\langle x_n,y_n\rangle\to 0$. Also, what do you mean exactly by "orthonormal sequences " ? For me it means that $\langle x_n,x_m\rangle =0 $ if $m\neq n$ and $\langle x_n,x_n\rangle =1 $ so that your reasoning does not hold. $\endgroup$ – Surb Nov 30 '18 at 12:53
  • $\begingroup$ Thank you, that's correct. Strange fact is that we are told to assume that the limit is $1$ $\endgroup$ – sound wave Nov 30 '18 at 12:58
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    $\begingroup$ That's strange, but nontrivial examples do exist, say the sine/cosine waves and their sawtooth wave analogy with the same frequencies. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 30 '18 at 13:03
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The two given bases are orthonormal, so $\langle x_n,x_n \rangle= \langle y_n,y_n \rangle =1$ for each $n$. Plug this back to your calculations.

\begin{align*} ||x_n − y_n||_2^2 &= \langle x_n − y_n,x_n − y_n \rangle \\ &= \langle x_n,x_n \rangle - \langle x_n,y_n \rangle - \langle y_n,x_n \rangle + \langle y_n,y_n \rangle\\ &= 2- 2\langle x_n,y_n \rangle \\ &\to 2-2(1)=0 \end{align*}

This gives the desired result.

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    $\begingroup$ How do you prove that $\langle x_n,y_n\rangle \to 1$? $\endgroup$ – Surb Nov 30 '18 at 12:57
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    $\begingroup$ @Surb That's an assumption in the question. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 30 '18 at 12:58
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    $\begingroup$ Thank you, I thought that $\langle x_n,x_n \rangle = 0$ because I see the sequence $\{x_n\}$ as a sequence of real numbers, and so each $x_i$ as a real number. But actually $x_i$ is a sequence and $\{x_n\}$ a sequence of sequences? So in this last case it will be $\langle x_n,x_n \rangle = ||x_n|| =1$ because the sequence $x_n$ is orthonormal? $\endgroup$ – sound wave Nov 30 '18 at 13:04
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    $\begingroup$ Ups it is $\langle x_n,x_n \rangle = ||x_n||^2$ $\endgroup$ – sound wave Nov 30 '18 at 13:11
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    $\begingroup$ @soundwave The question only suggests that $H$ is a Hibert space (i.e. a complete inner product space). It can be a set of numbers/functions/anything that satisfies the definition. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 30 '18 at 13:14

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