Uniform Convergence and Uniform Norm

Question

Given that the derivative of a given sequence of functions =
$$\frac{x}{x^2+\frac{1}{n}}$$
Show that it converges pointwise but not uniformly. I have that the function will converge to $\frac{x}{x^2}$ pointwise. However, I'm trying to show that it does not converge uniformy by using the uniform norm. I have it simplified into the following:
$$\frac{1}{n}\left\lVert\frac{x}{x^4+x^2\frac{1}{n}}\right\rVert_\infty$$ and I know I have to find the supremum of $$\left\lVert\frac{x}{x^4+x^2\frac{1}{n}}\right \rVert_\infty$$ but this is where I am running into trouble. I am not sure I fully understand how to determine what the supremum is in problems such as this.

Answer 1

You have not specified the domain of your functions, but I will assume that it is $\mathbb{R}$. You have the sequence of numbers $$\left\|\frac{x}{x^2} - \frac{x}{x^2 + \frac1n}\right\| = \frac1n\left\|\frac{x}{x^2 + \frac1nx^4}\right\|$$ which you would like to bound above zero in order to show that your sequence does not converge uniformly. (Here $\|\cdot\|$ is the sup norm over $\mathbb{R}$.)

By evaluating $x$ at any point, say $1/n$, we obtain a lower bound for this sup norm, say: $$\sup_x\left[\frac1n\left\|\frac{x}{x^2 + \frac1nx^4}\right\|\right] \geq \frac1n\left\|\frac{1/n}{1/n^2 + \frac1n1/n^4}\right\| = \left\|\frac{1}{1 + 1/n^3}\right\| \geq 0.5$$ whenever $n>0$. This is sufficient to demonstrate that the sequence does not converge uniformly.

Answer 2

Its simpler if you use the property of uniform convergence and differentiability of sequence of functions. We know that $$f_n'(x) = \dfrac{x}{x^2+1/n}$$ converges pointwise to $$\phi(x) = \dfrac{1}{x}$$ Suppose that $f_n'(x) \rightarrow \phi(x)$ uniformly, then we know that $$\phi'(x)=\lim_{n\to \infty}{f_n'(x)}\quad \forall x$$ has to be true. The above equation clearly fails for the given derivative sequence.