Here is Prob. 17, Sec. 6.1, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $f \colon I \to \mathbb{R}$ be differentiable at $c \in I$. Establish the Straddle Lemma. Give $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $u, v \in I$ satisfy $c-\delta(\varepsilon)<u\leq c \leq v < c+\delta(\varepsilon)$, then we have $\left\lvert f(v) - f(u) - (v-u)f^\prime(c) \right\rvert \leq \varepsilon (v-u)$. [Hint: The $\delta(\varepsilon)$ is given by Definition 6.1.1. Subtract and add the term $f(c) - c f^\prime(c)$ on the left side and use the Triangle Inequality.]
Here is Definition 6.1.1 in Bartle & Sherbert, 4th edition:
Let $I \subset \mathbb{R}$ be an interval, let $f \colon I \to \mathbb{R}$, and let $c \in I$. We say that a real number $L$ is the derivative of $f$ at $c$ if given any $\varepsilon > 0$ there exists $\delta(\varepsilon)>0$ such that if $x \in I$ satisfies $0 < \lvert x-c \rvert < \delta(\varepsilon)$, then $$ \tag{1} \left\lvert \frac{ f(x) - f(c) }{ x-c} - L \right\rvert < \varepsilon. $$ In this case we say that $f$ is differentiable at $c$, and we write $f^\prime(c)$ for $L$.
In other words, the derivative of $f$ at $c$ is given by the limit $$ \tag{2} f^\prime(c) = \lim_{x \to c} \frac{ f(x) - f(c) }{ x-c} $$ provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
My Attempt:
As $f$ is differentiable at $c \in I$, so there is a real number $f^\prime(c)$ such that given any real number $\varepsilon > 0$ we can find a real number $\delta(\varepsilon) > 0$ such that $$ \left\lvert \frac{ f(x) - f(c) }{ x-c} - f^\prime(c) \right\rvert < \varepsilon \tag{1} $$ or $$ \left\lvert \frac{ f(x) - f(c) - (x-c) f^\prime(c) }{ x-c} \right\rvert < \varepsilon \tag{1'} $$ for all $x \in I$ which satisfy $$ 0 < \lvert x-c\rvert < \delta(\varepsilon), $$ which is equivalent to $$ c-\delta(\varepsilon) < x < c+ \delta(\varepsilon) \ \mbox{ and } \ x \neq c. $$
So if $x \in I$ and $0 < \lvert x-c \rvert < \delta(\varepsilon)$, then upon multiplying both sides of (1') by $\lvert x-c\rvert$, we get $$ \left\lvert f(x)-f(c) - (x-c)f^\prime(c) \right\rvert < \varepsilon \lvert x-c \rvert. \tag{2} $$ And for $x=c$ both sides of (2) equal $0$. Therefore we can conclude that $$ \left\lvert f(x)-f(c) - (x-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert x - c \rvert \tag{2'} $$ for all $x \in I$ for which $c-\delta(\varepsilon) < x < c+\delta(\varepsilon)$.
From (2') we conclude that if $u, v \in I$ and $c-\delta(\varepsilon) < u \leq c \leq v < c + \delta(\varepsilon)$, then we have $$ \left\lvert f(u)-f(c) - (u-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert u-c \rvert = \varepsilon ( c-u ) \tag{2*} $$ and also $$ \left\lvert f(v)-f(c) - (v-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert v-c \rvert = \varepsilon (v-c), \tag{2**} $$ and therefore $$ \begin{align} & \ \ \ \ \left\lvert f(v) - f(u) - (v-u) f^\prime(c) \right\rvert \\ &= \left\lvert f(v) - f(c) + f(c) - f(u) - (v-c + c - u) f^\prime(c) \right\rvert \\ &= \left\lvert \left( f(v)-f(c) - (v-c)f^\prime(c) \right) + \left( f(c) - f(u) - (c-u)f^\prime(c) \right) \right\rvert \\ &\leq \left\lvert f(v)-f(c) - (v-c)f^\prime(c) \right\rvert + \left\lvert f(c) - f(u) - (c-u)f^\prime(c) \right\rvert \\ &\leq \varepsilon ( v-c ) + \varepsilon ( c-u ) \qquad \mbox{[ using (2*) and (2**) above ] } \\ &= \varepsilon (v-u). \end{align} $$
Is this proof good enough? Or, are there any problems in it?
Where has this lemma originated? Any applications of this lemma? Some references please.
