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I'm trying to understand one part of the fundamental theorem of calculus in the Spivak's calculus book. On page 282 he defined:

$$m_h=\inf\{f(x);c\le x\le c+h\}$$

$$M_h=\sup\{f(x);c\le x\le c+h\}$$

Afterwards, he said

$\lim_{h\to0}m_h=\lim_{h\to 0} M_h=f(c)$, since $f$ is continuous.

I understand intuitively why this is true, but I don't know how to prove it rigorously.

Community
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user42912
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2 Answers2

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Since $f$ is continuous, there is $x_h \in [c,c+h]$ such that $m_h=f(x_h)$.

With $h \to0$, we get $x_h \to c$. Then, by continuity, we derive $m_h \to f(c)$.

Fred
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Suppose that we create new functions $m(h)=\inf\{f(y):y \in [c-h,c+h]\}$ and $M(h)=\sup\{f(y):y \in [c-h,c+h]\}$. Prove that $\displaystyle \lim_{h \to 0}m(h)=f(c) \text{ and } \lim_{h \to 0}M(h)=f(c)$. This amounts to showing:

For an arbitrary $\varepsilon \gt 0$, can we find a $\delta_m$ and $\delta_M$ such that:

  1. $\forall h: 0\leq h\lt \delta_m:|m(h)-f(c)|\lt \varepsilon$

  2. $\forall h: 0 \leq h\lt \delta_M:|M(h)-f(c)|\lt \varepsilon$

Note that $m(0)$ and $M(0)$ trivially equal $f(c)$.

Proof 1

By assumption, we know that $f$ is continuous at $c \iff \displaystyle \lim_{k \to 0}:f(c+k)=f(c)$. This means there exists a $\delta_1: \forall k: |k| \lt \delta_1 \rightarrow |f(c+k)-f(c)| \lt\varepsilon.$ This implies that $f(c)-\varepsilon \lt f(c+k) \lt f(c)+\varepsilon$. Therefore, $f(c)-\varepsilon$ is a lower bound for $f(c+k)$. As such, $\forall k: |k| \lt \delta_1:f(c+k) \gt f(c)-\varepsilon$. This means that for all $x \in (c-\delta_1,c+\delta_1):f(x) \gt f(c) - \varepsilon$. Consider $0\lt d \lt \delta_1$. Necessarily, we have that $[c-d,c+d]\subset (c-\delta_1,c+\delta_1)$. Therefore, for any of these $d$s, we have that $ x \in [c-d,c+d] \implies f(x) \gt f(c)-\varepsilon$. Of course, then, we have that $m(d)\gt f(c)-\varepsilon$. By definition of infimum, we know that $m(d) \leq f(c)$. Therefore we immediately know that $m(d)-f(c) \leq 0 \implies m(d)-f(c) \lt \varepsilon$. This means that for all $\forall d: 0 \leq d \lt \delta_1\rightarrow|m(d)-f(c)| \lt \varepsilon$.

Proof 2

A similar proof can be used to show that there is a $\delta_M$ such that $\forall h: 0 \leq h\lt \delta_M:|M(h)-f(c)|\lt \varepsilon$. In fact, we can use the same $\delta_1$ as in Proof 1. Note that this will result in creating an upper bound for $f(x): f(x) \lt f(c)+\varepsilon$. Therefore, the supremum for any $x \in [c-d,c+d]$ where $d \lt \delta_1$ is also less than $f(c)+\varepsilon$. Or, equivalently, $M(d)-f(c) \lt \varepsilon$. Note that by definition of supremum, for any $d \in [0,\delta_1)$, we necessarily have that $M(d) \geq f(c) \implies M(d)-f(c) \geq 0 \implies M(d)-f(c) \gt -\varepsilon$. Therefore, we can conclude that: $\forall d: 0 \leq d \lt \delta_1 \rightarrow |M(d)-f(c)|\lt \varepsilon$

S.C.
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