# In the real domain, are there any theorems or definitions that state all functions are differentiable? [closed]

We know that not every function can be integrable, but as far as I know all functions are differentiable in the real domain.

My question: Are there any theorems or definitions that state that all functions are differentiable in real domain?

If there is any, can you state it or prove it?

Thanks.

## closed as unclear what you're asking by user147263, user230715, Milo Brandt, akusaja, Yagna PatelAug 21 '15 at 19:07

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• It's not true that all functions are differentiable. I'm not sure what you mean with "in the real domain". – Wojowu Aug 21 '15 at 16:11
• why should it? Is the absolute value differentiable at $0$? – user251257 Aug 21 '15 at 16:12
• What do you mean by "the real domain"? In standard mathematics, there are many functions $\mathbb R\to\mathbb R$ which are not differentiable. For example Dirichlet's function $$f(x) = \begin{cases} 1 & \text{if }x\in\mathbb Q \\ 0 & \text{otherwise} \end{cases}$$ – Henning Makholm Aug 21 '15 at 16:12
• @akusaja: None of the above counterexamples to your claim use complex numbers. – Henning Makholm Aug 21 '15 at 16:15
• @akusaja Functions can even be nowhere continuous! – egreg Aug 21 '15 at 16:17

When I arrived at university, my professor of mathematical analysis (twenty years ago, in Italy, the graduate program in mathematics used to have no calculus course at all, but directly mathematical analysis; we used Rudin's book) told us that the generic high school student believes that every function is of class $C^\infty$. Of course now you know that there are so many singular (i.e. non-differentiable) functions around you, but nevertheless I want to tell you the the most famous mathematicians who lived two centuries ago did believe that every function had a derivative.

The reason is that they - and probably you- thought that functions were elementary formulae like polynomials, trigonometric functions, logarithms and so on. These elementary functions are differentiable (up to some really unnatural cases) at all points of their domains of definition. Unluckily calculus courses teach us to deal with functions like everything was allowed: differentiating, integrating, finding inverses, etc.

So no, you can't prove in any way that any function is differentiable because that would be a wrong theorem.

• "the most famous mathematicians who lived two centuries ago did believe that every function had a derivative." This is a bit strange. Maybe "was differentiable almost everywhere" would be better? I'm thinking about examples like $f(x) = |x|$, as I don't know whether they were thought as functions back then. – Ian Mateus Aug 21 '15 at 16:35
• Well, in the 18th century a function used to be an explicit formula. The concept of zero measure was yet to come ;-) Of course an explicit formula can have singularities, but these points were always isolated and could be discarded for any purpose. For example $x \mapsto \sqrt{1-x^2}$ is singular at $x=\pm 1$, but I suspect that mathematicians were not scared by this fact. They did not think of a function as an intrinsic mathematical object: for this we had to wait until the beginning of the last century. – Siminore Aug 21 '15 at 16:41
• I was told, as a student, that at the beginning a function defined piecewise was usually rejected as a "wrong" object. The absolute value is familiar to us, but is a rather "suspicious" formula for an ancient mathematician: $|x|=\sqrt{x^2}$. – Siminore Aug 21 '15 at 16:48

Not every function is differentiable. Say, $f(x) = 0$ if $x<=0$ and $f(x) = 1$ if $x>0$. $f$ isn't differentiable at 0. Or, take a look at the Weierstrass function, which is continuous everywhere and differentiable nowhere.

Funny point of view:

We know that not every function can be integrable, but as far as I know all functions is differentiable in the real domain.

You know that not every function is integrable but you think that every function is differentiable? Have you every tried to differentiate a non-integrable function? You won't find any, because differentiability implies continuity implies integrability (at least on a compact domain).

So to find a function which is not differentiable, just take a function which is not integrable.

• So, if I've found the functions that can't be integrable, then we can say that the functions isn't differentiable too? – akusaja Aug 21 '15 at 16:43
• @akusaja On a compact domain that is true. If your function is defined on $\mathbb R$ and is not integrable that might also have other reasons (take for example $f(x)=x$). But I guess what you want is $\int_a^b g(x) dx$ for some real $a,b$. In that case your function $g$ has only to be defined on the compact interval $[a,b]$ and everything I said is applicable. – principal-ideal-domain Aug 21 '15 at 16:56