How to show that $f_{n}(x)=\frac{x^{2}}{x^{2}+(1-nx)^{2}},\,0\leq x\leq1 $ is not uniformly convergent

Question

The problem says:

$f_{n}(x)=\frac{x^{2}}{x^{2}+(1-nx)^{2}},\,0\leq x\leq1$ is convergent pointwise to $0$ but not uniformly.

So, I show that $f_n \rightarrow f$(pointwise) on $[0,1]$.

And I tried:

If I take $\epsilon=1/2N$, then

$$\frac{1}{2N}\leq\frac{1}{N}\left|\frac{1}{2-Nx}\right|=\left|\frac{x^{2}}{2Nx-N^{2}x^{2}}\right|=\left|\frac{x^{2}}{1-(1-Nx)^{2}}\right|\leq\left|\frac{x^{2}}{x^{2}-(1-Nx)^{2}}\right|$$ Thus, it is not convergent to $0$ uniformly on $[0,1]$.

But it seems to be wrong since $\epsilon$ is dependent on N.

How can I fix it?

Answer 1

Use the following propisition: "If $(f_n(x))$ converges pointwisley on A to f(x), and there is a sequence $(x_m)_{m\ge 1}$ in A such that $f_n(x_n)-f(x_n)$ does not converges to 0, then the given sequence of functions does not converges uniformly over A".

Now, take $x_n= \frac{1}{n}$.

Answer 2

Hint:

Try finding a suitable $N(\epsilon)$ by rearranging as $f_n(x) = \frac{1}{1 + (n-1/x)^2}$.

For non-uniform convergence consider $f_n(1/n)$

Answer 3

Note that $f_n(x)$ attains its maximum value when $nx=1\implies x=\dfrac{1}{n}$.

Also for $f_n\to f$ uniformly $\sup_{x\in [0,1]} |f_n(x)|\to 0$ which is false here as $\sup_{x\in [0,1]} |f_n(x)|=1$

Answer 4

To prove that $f$ is not uniformly convergent, you need to show that there is an $\epsilon >0$ such that for each $N\in \mathbb N$, there is an $n>N$ and $x_n$ with the property that $\vert f_n(x_n)-f(x_n)\vert >\epsilon$.

So, if we take $\epsilon =1/2$, and $x_n=1/n$, we have $f_n(x_n)=1$ and $f(x_n)=0$ then $\vert f_n(x_n)-f(x_n)\vert =1>\epsilon$ and the claim follows.

An easier way is to note that if $\left \{ f_n \right \}_{n}$ converges uniformly, then $\sup _{x\in \mathbb R}\vert f_n(x)-f(x)\vert\rightarrow 0$, so a routine calculus exercise on max/mins applied to $f_n$ leads again to $x_n=1/n$, from which we get an immediate contradiction.