"Greatest lower bound function"

Question

If $f $ is a function continuous at $c, h $ is positive and $m$ is a function defined as $ m(h)=\inf \{ f(x): x \in [c,c+h] \}$ , how can I prove that the limit of $ m $ as $ h $ approaches $ 0 $ is $ f(c)$ ?

Thanks

Answer 1

Hint : $\lim_{h \to 0}m(x) = \lim_{h \to 0}\inf \{ f(x): x \in [c,c+h] \}=\inf \lim_{h \to 0}\{ f(x): x \in [c,c+h] \}$(since $f$ is continuous )

Answer 2

Since $f$ is continuous at $c$, given $\varepsilon > 0$, there exists an $\delta > 0$ such that for all $x$, $x \in (c - \delta, c + \delta)$ implies $f(x) \in (f(c) - \varepsilon, f(c) + \epsilon)$. Let $h$ be a real number such that $0 < h < \delta$. Since $f(x) > f(c) - \varepsilon$ for all $x \in [c,c+h]$, $m(h) \ge f(c) - \varepsilon$. Also, $m(h) \le f(x) < f(c) + \varepsilon$ for all $x \in [c,c+h]$. Thus $|m(h) - f(c)| \le \varepsilon$ whenever $0 < h < \delta$. Consequently, $\lim_{h\to 0} m(h) = f(c)$.