8

Why does differentiability imply continuity, but continuity does not mean differentiability?

I am more interested in the part about a continuous function not being differentiable.

All I could find regarding why continuous functions can not be differentiable were counter-examples.

I just wanted to know if there was a more detailed explanation.

  • 7
    Because there are counterexamples. Such as the absolute value function. – Steven Gubkin Dec 02 '14 at 17:54
  • 4
    Because it's not true? For example $f(x)=|x|$ or $g(x)=x^{2/3}$, etc., etc. – JohnD Dec 02 '14 at 17:54
  • 1
    Differentiable means no corners and no jumps in particular. Continuous can have corners but not jumps. Both conditions are local - it does not have to be all corners (though it can be...) If it has corners (like the example given when x = 0) it cannot be differentiable at that corner. Note that Measurable can have corners and jumps. – Paul Dec 02 '14 at 18:00
  • 1
    Your question would be better received if it showed prior research. Did you search for "continuous function not differentiable"? The very first result is likely to be a Wikipedia page with an example. –  Dec 02 '14 at 18:35

3 Answers3

15

Here is an intuitive explanation.

Continuity requires that $f(x)-f(y)\to 0$ as $x - y \to 0$.

Differentiability requires that $f(x)-f(y)\to 0$ as $x - y \to 0$, and that $f(x)-f(y)\to 0$ at least as fast as $x - y \to 0$ (in the sense that the ratio still has a limit).

In particular, the conditions for differentiability include the condition for continuity.

MPW
  • 43,638
9

Why does differentiability imply continuity, but continuity does not imply differentiability?

Why are all differentiable functions continuous, but not all continuous functions differentiable? For the same reason that all cats are animals, but not all animals are cats. Differentiability also implies a certain “smoothness”, apart from mere continuity. It is perfectly possible for a line to be unbroken without also being smooth. Fractals, for instance, are quite “rugged” $($see first sentence of the third paragraph: “As mathematical equations, fractals are usually nowhere differentiable”$)$.

Lucian
  • 48,334
  • 2
  • 83
  • 154
2

Differentiability guarantees continuity in the following way :

Consider a function $f:R \rightarrow R$ which is differentiable at a point $x_0 \in R$ and it’s derivative is $L$. Thus by first principle derivative we have the RHD as

$$\lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h} = L$$ $$\implies \lim_{h\to0} [f(x_0 + h) - f(x_0)] = \lim_{h\to0} hL$$ $$\implies \lim_{h\to0} [f(x_0 + h) - f(x_0)]= 0$$

Which is the condition of continuity on the right hand side

We can prove the same for left hand side as well

Therefore differentiability always guarantees continuity but not the vice versa since the LHD May not be = RHD when trying to reverse engineer the above

A.Nag
  • 29