Continuous not implies differentiable - Revision history

Vipul: Created page with "{{function property non-implication| stronger = continuous function| weaker = differentiable function}} ==Statement== It is possible to have a function $f: \R \to \R$
2021-09-25T01:42:55Z

Created page with "{{function property non-implication| stronger = continuous function| weaker = differentiable function}} ==Statement== It is possible to have a function <math>f: \R \to \R</m..."

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{{function property non-implication|
stronger = continuous function|
weaker = differentiable function}}

==Statement==

It is possible to have a function <math>f: \R \to \R</math> that is a [[continuous function]] everywhere but is ''not'' a [[differentiable function]] everywhere, i.e., it has points where it is not differentiable.

Further, it is possible to construct an example where the function does not have a well-defined left-hand derivative or right-hand derivative.

==Proof==

===Example where one-sided derivatives exist and are not equal===

A [[piecewise definition by interval]] can be used to construct examples (see [[differentiation rule for piecewise definition by interval]]). The simplest example is the [[absolute value function]]:

<math>|x| := \left\lbrace \begin{array}{rl}x, & x \ge 0 \\ -x, & x < 0 \\\end{array} \right.</math>

[[File:Absolutevalue.png|500px]]

The function is continuous at 0. The left-hand derivative at 0 is -1 and the right-hand derivative is 1. These are not equal, so the function is not differentiable at 0.

Intuitively, the function is continuous at the point 0 where its definition changes, but it is not differentiable because it turns sharply (from moving down to moving up). There is no well-defined tangent line at (0,0) where it turns.

===Example where no one-sided derivatives exist===

Consider the function:

<math>g(x) := \left\lbrace\begin{array}{rl} x \sin(1/x), & x \ne 0 \\ 0, & x = 0 \\\end{array}\right.</math>

The expression for the function is continuous and differentiable for <math>x \ne 0</math>. The expression does not extend to <math>x = 0</math>, so we need to calculate the limit at 0 manually to determine if it is continuous and differentiable.

=== Proof of continuity at 0 ===

<math>\lim_{x \to 0} g(x) = \lim_{x \to 0} x \sin (1/x)</math>

We have a bound on absolute value:

<math>0 \le |x \sin (1/x)| = |x||\sin(1/x)| \le |x|</math>

Thus:

<math>-|x| \le x\sin (1/x) \le |x|</math>

Both the leftmost and rightmost expressions approach 0 as <math>x \to 0</math>, so by the [[sandwich theorem]], we get:

<math>\lim_{x \to 0} x \sin (1/x) = 0</math>

=== Proof of non-differentiability at 0 ===

The derivative at zero is the limit:

<math>\lim_{x \to 0} \frac{x \sin (1/x) - 0}{x - 0}</math>

This works out to:

<math>\lim_{x \to 0} \sin(1/x)</math>

The [[sine of reciprocal function]] does not have a limit as <math>x \to 0</math>, so this expression is not defined. In fact, even restricting to a one-sided limit gives an undefined expression (for more on this, see the explanation at [[intermediate value property not implies continuous]]).

== Related facts ==

* [[Intermediate value property not implies continuous]]: This uses the [[sine of reciprocal function]] directly, i.e., just <math>\sin(1/x)</math>.
* [[Derivative of differentiable function need not be continuous]]: This uses <math>x^2 \sin (1/x)</math>.