If $f$ is continous at $c$, prove $\lim_{h \to 0} (\inf \,\{f(x)\mid c \leqslant x \leqslant c+h\})=f(c)$

Question

Let $f$ be continous at $c$. Prove $$\lim_{h \to 0} \left(\inf \,\{f(x)\mid c \leqslant x \leqslant c+h\}\right)=f(c)$$

This fact is used in Spivak's book to prove 1nd Fundamental Calculus Theorem.

Answer 1

$\bigg|\lim_h \ \inf\ \bigg\{f(x)\bigg|c\leq x\leq c+h\bigg\} - f(c)\bigg|=\varepsilon >0 $

So there is $\delta$ s.t. for $|c-x|\leq \delta$, we have $|f(c)- f(x)|\leq \varepsilon/2$, by continuity.

When $h<\delta$ and $c\leq x\leq c+h$, then $$f(c)-\varepsilon/2\leq f(x) \leq f(c)+\varepsilon/2 $$

Answer 2

Let $\epsilon>0$. By continuity at $c$, there exists $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ for all $x$ with $|x-c|<\delta$. In particular, for any $h$ with $0<h<\delta$, we have $$f(c)-\epsilon<f(x)<f(c)+\epsilon$$ for all $x$ with $c\le x\le c+h$, hence $f(c)\ge \inf\{\,f(x)\mid c\le x\le c+h\,\}\ge f(c)-\epsilon$ for such $h$. Conclude.

Answer 3

Hopefully adding a bit of detail:

Hagen von Eizen:

$\epsilon \gt 0$ given.

For $0 < h < \delta$

$f(c)-\epsilon <f(x) < f(c)+\epsilon$,

for $x$ with $c \le x \le c+h$:

$f(c) \ge \inf$ { $f(x)| c \le x \le c+h$ } $\ge f(c)-\epsilon$ .

Consider sequences $\epsilon_n \rightarrow 0^+$, and $h_m \rightarrow 0^+$.

Continuity of $f$ at $c$:

For $\epsilon_n $ there is a $\delta_n >0$.

Let $h_{m_n}$ be a subsequence s.t. $0 < h_{m_n} \lt \delta_n.$

We now have:

$f(c) \ge \inf$ { $f(x)| c \le x \le c+h_{m_n}$ } $\ge f(c)-\epsilon_n$.

Take the limit $n \rightarrow \infty$.

Answer 4

You should note that the limit in question is based on $h\to 0^+$. Let $\epsilon >0$ be given and then there is a $\delta>0$ such that $$f(c) - \epsilon<f(x) <f(c) +\epsilon$$ whenever $c\leq x<c+\delta$. Now the function $g$ defined by $$g(h) =\inf\, \{f(x) \mid x\in[c, c+h] \} $$ is non-increasing in $(0,\delta)$ and is clearly bounded above by $f(c)$. It follows that the limit in question $\lim_{h\to 0^+}g(h)$ exists and lies in $[f(c) - \epsilon, f(c)] $. Since $\epsilon$ is arbitrary it follows that the limit is $f(c) $.