Let \(f:\mathbb{R}\to\mathbb{R}\) be defined by \(f(x)=sin(x^3)\). Then f is continuous but not uniformly continuous.

Hint: Can you find a sequence whose terms can get arbitrarily close to each other but the function gives distant values?


The function sin takes the value 1 at \(4n+1\) multiples of \(\pi/2\) and it is -1 at \(4n-1\) multiples of \(\pi/2\) .

Let \(x_n=(n\pi+\pi/2)^{1/3}\) That is, the function takes the value +1 when n is even and it is -1 when n is odd.

Now \(x_{n+1}-x_n\) \(=\) \(\frac{( n+1 )\pi+\pi/2-(n\pi+\pi/2)}{\text{terms involving n}}\)

This gives that the two terms \(x_{n+1}\) and \(x_n\) are close to each other. Because, the limiting value of the difference is zero. (So if you give me any positive real number \(\delta\) I can find an n such that the difference of two consecutive terms is less than that (\delta\) )

And what happens to \(f(x_n)\)? It is +1 and -1 for two consecutive terms (or -1 and +1). Therefore, the difference \(|f(x_{n+1})-f(x_n)|\) is always 2.

In particular, if I give \(\epsilon=1\) then whatever \(\delta\) you produce I will select two consecutive terms in the above sequence \(x_n\) which has distance less than \(\delta\) and the difference of values of \(f\) would not be less than 1.

This proves that \(f\) is not uniformly continuous.

Remark: \(f\) is continuous because it is a composition of two continuous functions ( the sine function applied to the polynomial function \(x\to x^3\) ).