# Why $f(x^k) - f(x^{k+1}) \rightarrow 0$ for monotonically non-increasing sequence $\{f(x^k)\}$?

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 $$f(x^k) - f(x^{k+1}) \rightarrow 0$$?

Thanks!

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 $$f(\bar x)$$.
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 $$\bar x$$.