Uniform convergence of a sequence of functions 4

Prove that the sequence $$\left((nx)/(1+4n^2x^2)\right)_{n\in\mathbb N}$$ is not uniformly convergent on $$(-a,a)$$, where $$a > 0$$

My attempt:

$$\lim_{n\to\infty}(nx)/(1+4n^2x^2) = 0 = f(x)$$

Now, compute$$M_n = \sup_{-a

Finally, compute $$\lim_{n\to\infty} M_n$$

The theorem says that if $$\lim_{n\to\infty}$$ $$M_n$$ = $$0$$, then $$\bigl(f_n(x)\bigr)_{n\in\mathbb N}$$ is uniformly convergent. And the question is to prove that it is not. Any help please? What is the supremum of $$f_n(x)$$, i.e what is $$M_n$$?

• What do you get for your supremum? – George Dewhirst May 9 at 10:50
• @GeorgeDewhirst I dont know what is the supremum, I edited the question, can you review it? – dodo bc May 9 at 10:52
• Sure basically you need to differentiate the function and set equal to zero, and find the turning points. See if they lie in $[-a,a]$. I suspect that for $n$ large enough they will. The turning point is likely to be a maximum. Supremum means maximise the function on the given interval. Once you have the maximum, it might be clear that this max as a function of $n$ won't tend to $0$ – George Dewhirst May 9 at 10:58

You can compute $$\sup f_n$$ by the standard method of computing $$f_n'$$ and determining where it is equal to $$0$$. It turns out that $$f_n'(x)=0\iff x=\pm\frac1{2n}$$ and that $$f_n\left(\pm\frac1{2n}\right)=\pm\frac14$$. So, $$\sup\lvert f_n\rvert=\frac14$$, and therefore your sequence does not converge uniformly to the null function.
We have $$f_n(\frac{1}{n})=\frac{1}{5}$$ for all $$n$$. Hence $$M_n \ge \frac{1}{5}$$ for all $$n$$.