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Let $(x_n)_{n}\subseteq E$ and $(y_n)_{n}\subseteq E$ such that $\|x_n\|=\|y_n\|=1$, where $E$ is a complex inner product space.

Consider the following property $(P)$:

$(P)$: There exists a constant $c< 1$ such that $$|\langle x_n, y_n\rangle|\leq c< 1,\;\forall n\in \mathbb{N}.$$

If $(P)$ fails to hold, why there exists a strictly increasing function $\varphi: \mathbb{N}\to \mathbb{N}$ such that $$\displaystyle\lim_{n\longrightarrow\infty}|\langle x_{\varphi(n)}\; ,\;y_{\varphi(n)}\rangle|=1.$$

My attempt: If $(P)$ fails to hold, then $\forall\, c<1$, there exists $n_c\in \mathbb{N}$ such that $|\langle x_{n_c}\; ,\;y_{n_c}\rangle|> c$.

If $c=\frac{n}{n+1}<1$, then $|\langle x_{n_c}\; ,\;y_{n_c}\rangle|> \frac{n}{n+1}$. On the other hand by Cauchy-Shwartz inequality $$|\langle x_{n_c}\; ,\;y_{n_c}\rangle| \leq \|x_{n_c}\|\|y_{n_c}\|=1.$$

My problem is how to construct explicitly the function $\varphi$

Note that this problem is taken from this note (Theorem 7.4. page 21)

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It is not true: if there is $N$ such that $|\langle x_N, y_N\rangle| =1$ and $|\langle x_n, y_n\rangle| =0$ for all $n\neq N$, then $(P)$ fails but for any strictly increasing $\varphi$ we have $\lim_{n\to\infty}|\langle x_{\varphi(n)}, y_{\varphi(n)}\rangle| =0$.

However, we can construct $\varphi$ inductively if for all $c\in (0,1)$, there exist infinitely many $n$ such that $$ |\langle x_n, y_n\rangle| > c. $$ Assume we have constructed strictly increasing $\varphi(j)$ for $j\le n$ for which $|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle| > 1-\frac{1}{2^j}$ holds. Since there are infinitely many $k$ such that $$ |\langle x_k, y_k\rangle| >1-\frac{1}{2^{n+1}},\tag{*} $$ we can find $k>\varphi(n)$ such that $(*)$ holds. By letting $\varphi(n+1)=k$, we have $\varphi(j)$ for $j\le n+1$ such that $|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle| > 1-\frac{1}{2^j}$ holds. By induction, $\varphi$ can be defined on $\mathbb{N}$ and it follows $$ \lim_{j\to\infty}|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle|=1 $$ from the construction.

EDIT: In view of the lecture note, the correct version of $(P)$ is: for some $c\in (0,1)$, $$ |\langle x_n, y_n\rangle|\leq c< 1$$ for all sufficiently large $n$. If we negate this statement, it follows that for all $c\in (0,1)$, $$\langle x_n, y_n\rangle|>c$$ for infinitely many $n$. Hence, the above construction works and gives the existence of a subsequence $\varphi(n)$ such that $$\lim\limits_{n\to\infty}|\langle x_{\varphi(n)}, y_{\varphi(n)}\rangle|=1.$$

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  • $\begingroup$ Please see my edit. Thanks you. $\endgroup$
    – Schüler
    Jan 8 '19 at 11:36
  • $\begingroup$ Thank you for your answer but I don't understand where is my wrong in my question. Even in the note the author writes there exists a constant $K\in \mathbb{R}$ and not in $(0,1)$. Thanks. $\endgroup$
    – Schüler
    Jan 8 '19 at 19:27
  • $\begingroup$ That's not the point. The range of $K$ (or $c$ in my term) doesn't matter. Note that if $K\le 0$, then we can pick a larger $K'\in (0,1)$ and say that $(P)$ holds for $K'$. So the range of constant does not make any difference to whether the property $(P)$ holds or not. Moreover, $(P)$ cannot be true for $K<0$. What matters is the part where you say "$\forall n\in\mathbb{N}$". It should be stated "for all sufficiently large $n$" instead.. $\endgroup$ Jan 8 '19 at 19:33
  • $\begingroup$ Thank you for the details but what is the difference between for all $n$ and for all sufficiently large n? That is for $n$ large enough? $\endgroup$
    – Schüler
    Jan 8 '19 at 19:40
  • $\begingroup$ To be precise, "for all sufficiently large $n$" means that there exists $N\ge 1$ such that the property holds for all $n$ greater than $N$. As I said in my answer, just one $n$ such that $|\langle x_n,y_n\rangle|=1$ can fail the $\forall n$ $(P)$. But "for all sufficiently large $n$" $(P)$ is not influenced by any single or finite $n$. $\endgroup$ Jan 8 '19 at 19:47

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