1

Question:

Suppose $f'$ is continuous on $[a,b]$ and $\epsilon\gt0$.Prove that there exist $\delta\gt0$ such that $|\frac{f(t)-f(x)}{t-x}-f'(x)|\lt\epsilon$ whenever $0\lt|t-x|\lt\delta$,with $x$ and $t$ in $[a,b]$.

My question is how can I prove it without using The Mean Value Theorem.

G.t.g.h
  • 152
  • Use the definition of $f'$. – on1921379 Aug 20 '22 at 14:03
  • It is probably impossible, because Mean Value Theorem is basically the only tool that allows to conclude properties of a function from properties of the derivative. One chance would be to use the Newton-Leibniz theorem, but its proof requires Mean Value Theorem too. – Adayah Aug 20 '22 at 14:25
  • 2
    @Adayah I don't think so. Because $f'$ is continuous on compact set $[a, b]$, so it's finite at every point in this interval and the limit $\lim_{t \to x} \frac{f(t) - f(x)}{t - x}$ exists, and equals $f'(x)$. Therefore, by definition, for every $\epsilon > 0$ there exists $\delta > 0$ such that $0 < |t - x| < \delta$ implies $|f'(x) - \frac{f(t) - f(x)}{t - x}| < \epsilon$. Is there an issue? – on1921379 Aug 20 '22 at 18:17
  • @on1921379 your choice of $\delta>0$ basically depends on $x$, which is not allowed in question. – user284331 Aug 21 '22 at 00:40
  • 1
    Been a while, but isn't this just uniform continuity, which you get since $f'$ is continuous on a compact set? – Alan Aug 23 '22 at 19:40
  • @Alan It is not. If you think it is, try writing down an actual proof and if you succeed, I will be very interested to see it. – Adayah Aug 25 '22 at 14:06

2 Answers2

0

Suppose it were false. Then there exists $r>0$ and there exist sequences $(t_n)_n$ and $(x_n)_n$ in $[a,b]$ with $t_n\ne x_n$ and with

$t_n,x_n$ both converging to some $y\in [a,b],$ and such that $$(*)\quad \left|\frac {f(t_n)-f(x_n)}{t_n-x_n}-f'(x_n)\right|>r$$ for every $n.$

Now $$\frac {f(t_n)-f(x_n)}{t_n-x_n}=\frac {1}{t_n-x_n}\int_{x_n}^{t_n}f'(u)du$$ is bounded above by $\max_{u\in [x_n,t_n]}f'(u)$ and bounded below by $\min_{u\in [x_n,t_n]}f'(u)$. So by $(*)$ there exists $u_n\in [x_n,t_n]$ with $$|f'(u_n)-f'(x_n)|>r.$$

But $u_n$ and $x_n$ both converge to $y$ so $f'$ is not continuous at $y,$ a contradiction.

DanielWainfleet
  • 57,985
  • 1
    This is definitely a correct answer. For your information, this is actually Exercise 48.3 from the book Foundations Of Mathematical Analysis by Richard Johnsonbaugh and W. E. Pfaffenberger, but integration theory has not covered yet before that chapter, not even mean value theorem, I always think the authors have simply put the wrong order of the exercise. – user284331 Aug 24 '22 at 00:07
  • @user284331. Interesting. Note that the upper & lower bounds for the integral, and the continuity of $f'$, gives us the MVT for the case when $f'$ is continuous, which is used in the 2nd half of the proof of the Fundamental Theorem of Calculus. – DanielWainfleet Aug 24 '22 at 01:13
  • I think that instead of picking the sequences it is cleaner to just use the uniform continuity of $f'$ (which you basically prove) and proceed in the same way. – Adayah Aug 25 '22 at 14:28
0

I had the same idea as Adayah's suggestion in his comment (which I hadn't read before) of DanielWainfleet's answer.

Since $f'$ is continuous on $[a,b]$, it is uniformly continuous, hence there exists $\delta>0$ such that forall $u,x\in[a,b]$: $$|u-x|<\delta\Longrightarrow|f'(u)-f'(x)|<\epsilon.$$ Then, forall $x,t\in[a,b]$ such that $0<|t-x|<\delta$: $$\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|=\left|\frac1{t-x}\int_x^t(f'(u)-f'(x))\,\mathrm du\right|<\epsilon.$$

Anne Bauval
  • 34,650