Exploring the Conditions for Differentiability in Continuous Functions

The relationship between continuity and differentiability is a fundamental aspect of real analysis. While intuitively, it might seem that a function, which is continuous everywhere and differentiable at a particular point, must be differentiable everywhere, this is not always the case. In this article, we will delve into this misconception and provide a counterexample to demonstrate the subtleties involved in the relationship between these two properties.

Introduction to Continuity and Differentiability

Continuous and differentiable functions are central concepts in calculus. A function ( f: mathbb{R} to mathbb{R} ) is said to be continuous at a point ( x a ) if ( lim_{x to a} f(x) f(a) ). On the other hand, ( f ) is differentiable at ( x a ) if the derivative ( f'(a) ) exists, which means the limit ( lim_{h to 0} frac{f(a h) - f(a)}{h} ) exists.

A Myth Debunked: Continuous Everywhere, Differentiable Nowhere

There is a common misconception that if a function is continuous everywhere and differentiable at at least one point, then it must be differentiable everywhere. However, this is not true in general. We will illustrate this with a specific counterexample.

Counterexample: A Function Continuous Everywhere but Differentiable Nowhere Except at One Point

Consider the function ( f: mathbb{R} to mathbb{R} ) defined by ( f(x) x ). This function is continuous everywhere and differentiable everywhere. The derivative of ( f(x) ) is ( f'(x) 1 ) for all ( x in mathbb{R} ), and particularly at any point ( x a ), the derivative is clearly defined as ( f'(a) 1 ).

However, consider the function ( g(x) x^2 sinleft(frac{1}{x}right) ) for ( x eq 0 ) and ( g(0) 0 ). This function is continuous everywhere and differentiable at all points except at ( x 0 ).

Proof of Continuity and Differentiability:

Continuity at ( x 0 ): We need to show ( lim_{x to 0} g(x) g(0) ). Since ( -|x| leq x^2 sinleft(frac{1}{x}right) leq |x| ), we have ( lim_{x to 0} g(x) 0 g(0) ), so ( g(x) ) is continuous at ( x 0 ).

Differentiability at ( x 0 ): We need to check if the limit ( lim_{h to 0} frac{g(h) - g(0)}{h} ) exists. Evaluating the limit, we get:

[begin{align*}lim_{h to 0} frac{g(h) - g(0)}{h} lim_{h to 0} frac{h^2 sinleft(frac{1}{h}right)}{h} lim_{h to 0} h sinleft(frac{1}{h}right) 0.end{align*}]

The limit is zero, so ( g(x) ) is differentiable at ( x 0 ) and ( g'(0) 0 ).

Non-differentiability at other points: For ( x eq 0 ), the derivative of ( g(x) ) can be computed as:

[g'(x) 2x sinleft(frac{1}{x}right) - cosleft(frac{1}{x}right).]

As ( x to 0 ), the term ( 2x sinleft(frac{1}{x}right) ) tends to zero, but ( cosleft(frac{1}{x}right) ) oscillates between -1 and 1. Therefore, the derivative of ( g(x) ) oscillates and does not approach a limit, implying that ( g(x) ) is not differentiable at any ( x eq 0 ).

The Myth Revisited: Continuous Everywhere, Nowhere Differentiable Except at One Point

The counterexample provided above (i.e., ( f(x) x ) for all ( x )) is quite different from the myth of a continuous function that is differentiable at one point but not everywhere. The function ( g(x) x^2 sinleft(frac{1}{x}right) ) (for ( x eq 0 ), and 0 for ( x 0 )) illustrates a case where a continuous function is differentiable everywhere, except at a single point.

Conclusion

In summary, the relationship between continuity and differentiability is nuanced. While continuity at a point does not guarantee differentiability at that point, and vice versa, the misconception that continuity everywhere and differentiability at a single point implies everywhere differentiability does not hold. A function can be continuous everywhere and differentiable at all points except one, as demonstrated by the function ( g(x) x^2 sinleft(frac{1}{x}right) ).

Keywords

differentiability, continuous functions, theorem proof

References

Citations and references can be added here if needed, but for the sake of this article, a general overview has been provided.