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Suppose $f'$ is continuous on $[a,b]$ and $\epsilon >0$ is given. Prove that $\exists\; \delta>0$ such that $$\left\lvert \frac{f(t)-f(x)}{t-x}-f'(x)\right\rvert<\epsilon$$

$\forall \;0<|t-x|<\delta, a\le x \le b ,a\le t\le b$.

How to solve this problem since using compactness argument ?

Gibbs
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Understanding math
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2 Answers2

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Since $f'$ is continuous we may write $$f(t)-f(x)=\int_x^tf'(\xi)\>d\xi=(t-x)\int_0^1f'\bigl(x+\tau(t-x) \bigr)\>d\tau\ ,$$ and therefore $${f(t)-f(x)\over t-x}-f'(x)=\int_0^1\bigl(f'\bigl(x+\tau(t-x)\bigr) -f'(x)\bigr)\>d\tau\qquad(t\ne x)\ .\tag{1}$$ Now $f'$ is automatically uniformly continuous on $[a,b]$. Given an $\epsilon>0$ we therefore can find a $\delta>0$ such that $0<|h|<\delta$ implies $\bigl|f'(x+h) -f'(x)\bigr|<\epsilon$ for arbitrary $x$, $x+h\in[a,b]$. From $(1)$ it then follows that $$\left|{f(t)-f(x)\over t-x}-f'(x)\right|<\epsilon$$ whenever $0<|t-x|<\delta$, and both $x$ and $t$ are in $[a,b]$.

Christian Blatter
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Following from my hint I gave in the comments:

$f'$ is continuous on the compact interval $[a,b]$, and thus is uniformly continuous on $[a,b]$.

$\therefore \ \forall \epsilon>0$, $\exists \ \delta>0$, such that for all $x,t\in [a,b]$, we have

$$0<|x-t|<\delta\implies|f'(x)-f'(t)|<\epsilon$$

Because $f'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, by the mean value theorem, there exists some $c\in[x,t]$, such that

$$\frac {f(t)-f(x)}{t-x}=f'(c)$$

The claim then follows

$$\bigg| \frac {f(t)-f(x)}{t-x}-f'(t) \bigg|=|f'(c)-f'(t)|<\epsilon$$

For $0<|x-t|=|t-x|<\delta$

Mark Pineau
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