Prove that if $t:=\lim_n t_n >0$, then $(u_n)$ converges

Question

I'm trying to solve below exercise, i.e.,

Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(u_n) \subset H$ and $(t_n) \subset \mathbb R_{>0}$ such that $(t_n)$ is non-increasing. Assume that $$ \langle t_n u_n-t_m u_m, u_n-u_m \rangle \le 0 \quad \forall m, n \in \mathbb N. $$

1. Prove that either $|u_n| \to \infty$ or $(u_n)$ converges.
2. Prove that if $t:=\lim_n t_n >0$, then $(u_n)$ converges.

1. Could you verify my attempt on (1.)?
2. Could you elaborate on how to solve (2.)?

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We have $$ \begin{aligned} \langle t_{n+1} u_{n+1}-t_n u_n, u_{n+1}-u_n \rangle &\le 0 \\ \iff |u_{n+1}-u_n|^2 &\le \frac{t_n - t_{n+1}}{t_{n}} \langle u_{n+1}, u_{n+1}-u_{n} \rangle \\ \implies |u_{n+1}-u_n|^2 &\le \langle u_{n+1}, u_{n+1}-u_{n} \rangle \qquad (1) \\ \iff |u_{n+1}|^2 - 2 \langle u_n, u_{n+1} \rangle + |u_n|^2 &\le |u_{n+1}|^2 - \langle u_n, u_{n+1} \rangle \\ \iff |u_{n}|^2 &\le \langle u_n, u_{n+1} \rangle \qquad (2)\\ \implies |u_n| &\le |u_{n+1}|. \end{aligned} $$

So $(|u_n|)_n$ is non-decreasing.

1. Assume that $|u_n| \not\to \infty$. Then $(|u_n|)_n$ is bounded and convergent in $\mathbb R$. From $(1)$ and $(2)$, we have $$ \begin{aligned} |u_{n+1}-u_n|^2 &\le \langle u_{n+1}, u_{n+1}-u_{n} \rangle \\ \iff |u_{n+1}-u_n|^2 &\le |u_{n+1}|^2 - \langle u_n, u_{n+1} \rangle \\ \implies |u_{n+1}-u_n|^2 &\le |u_{n+1}|^2 - |u_n|^2. \\ \end{aligned} $$

Because $(|u_n|^2)_n$ is a Cauchy sequence, so is $(u_n)_n$.

Answer 1

Concerning $2,$ the inequality $$|u_{n+1}-u_n|^2 \le \frac{t_n - t_{n+1}}{t_{n}} \langle u_{n+1}, u_{n+1}-u_{n} \rangle$$ implies $$|u_{n+1}|-|u_n|\le |u_{n+1}-u_n|\le \frac{t_n - t_{n+1}}{t_{n}}|u_{n+1}|\le \frac{t_n - t_{n+1}}{t}|u_{n+1}|$$ Hence $$ \left (1- \frac{t_n - t_{n+1}}{t}\right )|u_{n+1}|\le |u_{n}|$$ Let $a_n=(t_{n}-t_{n+1})/t.$ Then $a_n<1$ for $n$ large enough. WLOG we may assume $a_n<1$ for all $n$ (by dropping finitely many terms of the sequence $u_n$ and renumerating the remaining terms). Thus we obtain $$|u_{n+1}|\le (1-a_n)^{-1}|u_{n}|$$ By iterating we get $$|u_{n+1}|\le \left (\prod_{k=1}^n(1-a_k)\right)^{-1}|u_1|$$ The partial products $\prod_{k=1}^n(1-a_k)$ are convergent to a positive number as the series $\sum_{k=1}^\infty a_k$ is convergent (to $(t_1-t)/t$). Hence the sequence $|u_n|$ is bounded. Now we can apply $1.$ to get convergence of $u_n.$

Answer 2

I have found a proof of (2.) as follows.

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We have $$ \begin{aligned} \langle t_n u_n-t_m u_m, u_n-u_m \rangle &\le 0 \\ \iff |u_n - u_m|^2 &\le \frac{t_n - t_m}{t_n} \langle u_m, u_m-u_n \rangle \\ \implies |u_n - u_m| &\le \frac{|t_n - t_m|}{t} |u_m|. \end{aligned} $$

Fix $\varepsilon >0$. Because $t_n \to t$, there is $\bar m \in \mathbb N$ such that $$ |t_n - t_m| \le \varepsilon \quad \forall m,n \ge \bar m. $$

Then $$ |u_n - u_{\bar m}| \le \frac{\varepsilon}{t} |u_{\bar m}| \quad \forall n \ge \bar m. $$

It follows that $(u_n)$ is bounded. The claim then follows from (1.)