Elementary proof that the derivative of a real function is continuous somewhere

One can use the Baire category theorem to show that if f:RR is differentiable, then f is continuous at some cR. Is there an elementary proof of this fact? By “elementary” I mean at the level of intro real analysis.

Edit: In spite of the decent response this question has gotten, after more than two and a half months there are still no answers. It’s perhaps possible that there’s some “deep” reason we should not expect an elementary proof of this. I will therefore also accept a well reasoned discussion as to why such a proof is unlikely.


One can actually prove with elementary tools something stronger, namely the following

Theorem: If f:RR is differentiable in a (non-degenerate) interval [a,b] then f is continuous at some point c(a,b) (a corollary being that if f is differentiable throughout R, then f if continuous on a dense subset of R).

Proof: For any [u,t][a,b], define osc(u,t)=sup (with osc(u,t)=+\infty if the right-hand side is unbounded; this can easily be made more formal with some more verbiage). Informally, osc(u,t) tells us how much the slope of f can “oscillate” in the interval [u,t]; the essence of the proof is showing that osc(u,t) must converge to 0 as u and t converge to some point c, and that c is then a point at which f’ is continuous.

Consider a generic sequence of “concentric and convergent” intervals [u_i,t_i], i.e. one that satisfies u_i\leq u_{i+1}<t_{i+1}\leq t_i and |t_i-u_i|\rightarrow 0. Let u_i,t_i\rightarrow v (note that both u_i and t_i are monotone and bounded, so the limit exists); then \lim \frac{f(t_i)-f(u_i)}{t_i-u_i}=f’(v), since f is differentiable in v and thus both \frac{f(t_i)-f(v)}{t_i-v} and \frac{f(v)-f(u_i)}{v-u_i} converge to f’(v), and so \frac{f(t_i)-f(u_i)}{t_i-u_i} which is a convex linear combination of \frac{f(t_i)-f(v)}{t_i-v} and \frac{f(v)-f(u_i)}{v-u_i} also converges to f’(v).

Let us now show that there is a sequence of concentric and convergent intervals [u_i,t_i]\subseteq (a,b) for which osc(u_i,t_i)\rightarrow 0. Suppose it were not the case. Then, starting with an arbitrary [w_0, z_0]\subseteq(a,b) there would be an \epsilon>0 such that given [w_i,z_i] we could always find a non-degenerate [w_{i+1},z_{i+1}]\subseteq [w_i,z_i] such that \left|\frac{f(z_i)-f(w_i)}{z_i-w_i} - \frac{f(z_{i+1})-f(w_{i+1})}{z_{i+1}-w_{i+1})}\right|>\epsilon (note that \epsilon is independent of i). Furthermore, such [w_i,z_i] could always be chosen arbitrarily small, since if g:\mathbb{R}\rightarrow\mathbb{R} is continuous in a non-degenerate interval [\alpha,\beta], then for any \delta>0 there exists a non-degenerate interval [\alpha',\beta']\subseteq(\alpha,\beta) with |\beta'-\alpha'|<\delta and \frac{g(\beta)-g(\alpha)}{\beta-\alpha}=\frac{g(\beta')-g(\alpha')}{\beta'-\alpha'} (the proof is essentially identical to that of Rolle's theorem, but stopping before taking the differentiation limit). Thus \frac{f(z_i)-f(w_i)}{z_i-w_i} would not converge to a (finite) limit, contradicting the differentiability of f in \lim w_i =
\lim z_i

Then, consider a sequence of concentric and convergent intervals [u_i,t_i]\subseteq(a,b), with u_i,t_i\rightarrow c, for which osc(u_i,t_i)\rightarrow 0. It is immediate to see that osc(u_i,t_i)=max\left(\left(\left(\sup_{[w,z]\subseteq[u_i,t_i]} \frac{f(z)-f(w)}{z-w}\right) - \frac{f(t_i)-f(u_i)}{t_i-u_i}\right), \left(\frac{f(t_i)-f(u_i)}{t_i-u_i} - \left(\inf_{[w,z]\subseteq[u_i,t_i]} \frac{f(z)-f(w)}{z-w}\right)\right)\right), so if osc(u_i,t_i)\rightarrow 0 then \sup_{[w,z]\subseteq[u_i,t_i]}\frac{f(z)-f(w)}{z-w}, \inf_{[w,z]\subseteq[u_i,t_i]
} \frac{f(z)-f(w)}{z-w}\rightarrow \lim \frac{f(t_i)-f(u_i)}{t_i-u_i}
= f’(c)
. And since in any interval [u_i,t_i] we have that \sup_{[w,z]\subseteq[u_i,t_i]}\frac{f(z)-f(w)}{z-w} \geq f' \geq \inf_{[w,z]\subseteq[u_i,t_i]
} \frac{f(z)-f(w)}{z-w}
, then f’(x)\rightarrow f’(c) as x\rightarrow c, i.e. f’ is continuous at c.

Source : Link , Question Author : MathematicsStudent1122 , Answer Author : Anonymous

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