# Construct a strictly increasing function $\varphi: \mathbb{N}\to \mathbb{N}$

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) 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 $$\lim\limits_{n\to\infty}|\langle x_{\varphi(n)}, y_{\varphi(n)}\rangle|=1.$$
• 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. Jan 8 '19 at 19:27
• 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.. Jan 8 '19 at 19:33
• 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? Jan 8 '19 at 19:40
• 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 $n$. Jan 8 '19 at 19:47