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

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.

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 $$0, we have $$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.

• I would use archimedian property and finish it, only one thing was not too clear for me: the conclusion after the "hence" – Jordan May 12 at 13:30

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

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 $$f(c)-\varepsilon/2\leq f(x) \leq f(c)+\varepsilon/2$$

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 $$f(c) - \epsilon whenever $$c\leq x. 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 $$f(c)$$.

Hopefully adding a bit of detail:

Hagen von Eizen:

$$\epsilon \gt 0$$ given.

For $$0 < h < \delta$$

$$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 $$n \rightarrow \infty$$.