It could be possible that "almost all" everywhere continuous functions are nowhere differentiable, but I'll have to give that one a bit more thought.
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Brian Charlesworth
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11 months, 2 weeks ago

If you, by the word 'everywhere', mean everywhere in its domain, then I can give you a much simpler function: Define the function \(f:N\rightarrow R\) as \(f\left( n \right)=n,\forall n\in N\). Clearly, since each point in its domain is isolated, we can say that it is continuous 'everywhere' in its domain by default, but not differentiable anywhere.
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Kuldeep Guha Mazumder
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9 months, 3 weeks ago

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@Kuldeep Guha Mazumder
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Good point. By most definitions of continuity a discrete function, (i.e., a function in which its domain is at most countable), it is considered "vacuously true" that the function is continuous at all of its (isolated) points. We could also then consider the function \(f:\mathbb{Q} \rightarrow \mathbb{R}, f(x) = x\) for all \(x \in \mathbb{Q}\) as an example.
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Brian Charlesworth
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9 months, 3 weeks ago

## Comments

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TopNewestThe Weierstrass function is the most commonly cited example. Here and here are other examples, not to mention fractals.

It could be possible that "almost all" everywhere continuous functions are nowhere differentiable, but I'll have to give that one a bit more thought. – Brian Charlesworth · 11 months, 2 weeks ago

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– Hobart Pao · 11 months, 2 weeks ago

Yup, that's what I was thinking of.Log in to reply

If you, by the word 'everywhere', mean everywhere in its domain, then I can give you a much simpler function: Define the function \(f:N\rightarrow R\) as \(f\left( n \right)=n,\forall n\in N\). Clearly, since each point in its domain is isolated, we can say that it is continuous 'everywhere' in its domain by default, but not differentiable anywhere. – Kuldeep Guha Mazumder · 9 months, 3 weeks ago

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– Brian Charlesworth · 9 months, 3 weeks ago

Good point. By most definitions of continuity a discrete function, (i.e., a function in which its domain is at most countable), it is considered "vacuously true" that the function is continuous at all of its (isolated) points. We could also then consider the function \(f:\mathbb{Q} \rightarrow \mathbb{R}, f(x) = x\) for all \(x \in \mathbb{Q}\) as an example.Log in to reply

– Kuldeep Guha Mazumder · 9 months, 3 weeks ago

Yes yes I was trying to point that out only.Log in to reply

Oh sorry I didn't see your post is already replied. – Ravi Dwivedi · 11 months, 2 weeks ago

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Weierstrass function is such a function – Ravi Dwivedi · 11 months, 2 weeks ago

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