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

One can use the Baire category theorem to show that if $f:\mathbb{R} \to \mathbb{R}$ is differentiable, then $f'$ is continuous at some $c \in \mathbb{R}$. 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.

## Answer

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

Theorem: If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable in a (non-degenerate) interval $[a,b]$ then $f'$ is continuous at some point $c\in(a,b)$ (a corollary being that if $f$ is differentiable throughout $\mathbb{R}$, then $f’$ if continuous on a dense subset of $\mathbb{R}$).

Proof: For any $[u,t]\subseteq[a,b]$, define $osc(u,t)=\sup_{[w,z]\subseteq[u,t]}\left|\frac{f(t)-f(u)}{t-u} - \frac{f(z)-f(w)}{z-w}\right|$ (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} 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 $c$.

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