I have been doing a lot of problems regarding calculus. An utmost basic question I stumble upon is "when is a continuous function differentiable?" (irrespective of whether its in an open or closed set).
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I don't understand what "irrespective of whether it is an open or closed set" means. What set? Where? – Qiaochu Yuan May 13 '15 at 17:33
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Differentiability implies a certain “smoothness” on top of continuity. For instance, we can have functions which are continuous, but “rugged”. – Lucian May 13 '15 at 21:52
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well try to see from my perspective its not exactly duplicate since i went through the Lagranges theorem where it says if every point within an interval is continuous and differentiable then it satisfies the conditions of the mean value theorem, note that it defines it for every interval same does the work cauchy's theorem and fermat's theorem that is they can be applied only to closed intervals so when i faced question for open interval i was forced to ask such a question – Arnav Das May 14 '15 at 11:57
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This is an old problem in the study of Calculus. Before the 1800s little thought was given to when a continuous function is differentiable. It was commonly believed that a continuous function is differentiable practically everywhere on its domain, except for a couple of obvious places, like the kink of the absolute value of $x$.
One obstacle of the times was the lack of a concrete definition of what a continuous function was. A formal definition, in the $\epsilon-\delta$ sense, did not appear until the works of Cauchy and Weierstrass in the late 1800s.
Weierstrass in particular enjoyed finding counter examples to commonly held beliefs in mathematics. His most famous example was of a function that is continuous, but nowhere differentiable: $$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$$ where $a \in (0,1)$, $b$ is an odd positive integer and $$ab > 1 + \frac32 \pi.$$
As an answer to your question, a general continuous function does not need to be differentiable anywhere, and differentiability is a special property in that sense.
On the other hand, if you have a function that is "absolutely" continuous (there is a particular definition of that elsewhere) then you have a function that is differentiable practically everywhere (or more precisely "almost everywhere").
Joel
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i faced a question like if F be a function upon all real numbers such that F(x) - F(y) <_(less than or equal to) C(x-y) where C is any real number for all x & y then F must be differentiable or continuous ? by Lagranges theorem should not it be differentiable and thus continuous rather than only continuous ? – Arnav Das May 13 '15 at 16:26
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1If $|F(x)-F(y)| < C |x-y|$ then you have only that $F$ is continuous. Take for instance $F(x) = |x|$ where $|F(x)-F(y)| = ||x|-|y|| < |x-y|$. $F$ is not differentiable at the origin. – Joel May 13 '15 at 16:52
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However, such functions are absolutely continuous, and so there are points for which they are differentiable. These functions are called Lipschitz continuous functions. – Joel May 13 '15 at 16:54
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Radamachers differentation theorem says that a Lipschitz continuous function $f:\mathbb{R}^n \mapsto \mathbb{R}$ is totally differentiable almost everywhere.
Beginning at page. 226 of An introduction to measure theory by Terence tao, this theorem is explained.
Henrik
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Other example of functions that are everywhere continuous and nowhere differentiable are those governed by stochastic differential equations. For example, let $X_t$ be governed by the process (i.e., the Stochastic Differential Equation)
$$dX_t=a(X_t,t)dt + b(X_t,t) dW_t \tag 1$$
where $W_t$ is a Wiener process and the functions $a$ and $b$ can be $C^{\infty}$. Then it can be shown that $X_t$ is everywhere continuous and nowhere differentiable.
To give an simple example for which we have a closed-form solution to $(1)$, let $a(X_t,t)=\alpha X_t$ and $b(X_t,t)=\beta X_t$. Then, using Ito's Lemma and integrating both sides from $t_0$ to $t$ reveals that
$$X_t=X_{t_0}e^{(\alpha-\beta^2/2)(t-t_0)+\beta(W_t-W_{t_0})}$$
The reason that $X_t$ is not differentiable is that heuristically, $dW_t \sim dt^{1/2}$. Thus, the term $dW_t/dt \sim 1/dt^{1/2}$ has no meaning and, again speaking heuristically only, would be infinite.
Inasmuch as we have examples of functions that are everywhere continuous and nowhere differentiable, we conclude that the property of continuity cannot generally be extended to the property of differentiability.
Mark Viola
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