Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and has a left derivative, $f^-$, everywhere in a neighborhood of $x.$ Suppose $f^-$ is continuous at $x.$ Does this imply that $f$ is differentiable at $x$?
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2No, let $f=\chi_{(0,\infty)}.$ Perhaps you want to assume $f$ is continuous at $x$? – zhw. Apr 05 '15 at 17:27
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Thanks. The function I am trying to apply this conjecture to is indeed continuous. – Mike Apr 05 '15 at 17:52
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See here : Continuous right derivative implies differentiability
This is almost exactly the same.
Mathematician 42
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The question you link to is very related, but it's not quite the same, as only continuity at one point $x$ is hypothesized. Can the solution be modified? – Olivier Bégassat Apr 05 '15 at 18:53
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@OlivierBégassat I don't think that his proof can be modified to show something like this. Continuity at one point isn't a very strong condition, you know. But on the other hand, I can't think of any counter examples. – Kitegi Apr 05 '15 at 18:59
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Should be true. Very closed to Continuous right derivative implies differentiability
Lemma Let $g$ be continuous in [a, b] and continuous left derivative in (a,b]. $g(a)=g(b)$. At the minimal $g(m)$ satisfies $g'(m-)\geqslant 0$ and the maximal $g(M)$ satisfies that $g'(M-)\leqslant 0$. Notice that both $m$ and $M$ belong to (a,b] because $g(a)=g(b)$.
Applying the lemma to $l(x)=f(x) - \dfrac{f(b)-f(a)}{b-a}(x-a)$ where $f$ as above. We have there are $m, M \in (a,b]$,$$l'(m-) = f'(m-) - \frac{f(b)-f(a)}{b-a} \geqslant 0$$ and$$ l'(M-) = f'(M-) - \frac{f(b)-f(a)}{b-a} \leqslant 0. $$ That is$$ \inf_{x\in(a,b]}f'(x-) \leqslant f'(M-) \leqslant \frac{f(b)-f(a)}{b-a} \leqslant f'(m-) \leqslant \sup_{x\in(a,b]}f'(x-).$$
Thus $$\lim_{x\to b+}f'(x-)= \lim\inf_{x\to b+}f'(x-)\leqslant \lim_{y\to x+} \frac{f(y)-f(x)}{y-x} \leqslant \lim\sup_{x\to b+} f'(x-) = \lim_{x\to b+} f'(x-)$$. Then it is obtained that $$f'(x+) = \lim_{x\to b+} f'(x-) = f'(x-).$$ Thus it is differentialable.
