Given $\{f(x^k)\}$ a monotonically non-increasing sequence $\{f(x^k)\}$ converges to a finite value or diverges to $-\infty$. Since $f$ is continuous, $f(\bar{x})$ is a limit point of $\{f(x^k)\}$, so it follows that the entire sequence $\{f(x^k)\}$ converges to $f(\bar{x})$ and $f(x^k) - f(x^{k+1}) \rightarrow 0$.

My questions are:

  1. Why $f(\bar{x})$ is a limit point of $\{f(x^k)\}$?
  2. Why the entire sequence $\{f(x^k)\}$ converges to $f(\bar{x})$?
  3. Why $f(x^k) - f(x^{k+1}) \rightarrow 0$?
  4. Given that $\{x^k\}$ converges to a stationary point $\bar{x}$ why $f(x^k) - f(x^{k+1}) \rightarrow 0$?



1) $(x_k)$ has a subsequence converging to $\bar x$. Since $f$ is continuous, the same subsequence now taken from $(f(x_k))$ converges to $f(\bar x)$.

2) This is a property of monotonic sequences.

3) If $f(x_k)\to f(\bar x)$, then also $f(x_{k+1})\to f(\bar x)$.

4) Follows from (1)-(3). Has nothing to do with stationarity of $\bar x$.

  • $\begingroup$ Thanks for answering. Could you please provide lite more details about the property mentioned in 2)? $\endgroup$
    – Thoth
    Jan 15 '20 at 13:20
  • $\begingroup$ math.stackexchange.com/questions/945328/… $\endgroup$
    – daw
    Jan 15 '20 at 14:57

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