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Define a sequence $(f_n)$ of functions on $\Bbb{R}$ by putting $f_n(x) := \frac{x}{1+nx^2}$ for each natural number $n$ and for each real $x$. Let $f(x)$ be the limit of the sequence $(f_n(x))$ and $g(x)$ be the limit of the sequence (f_n' (x)) (the derivative of $(f_n(x))$). Find $f$ and $g$. Find all intervals $I$ where the sequences are uniformly convergent. For what $x$ is f\,'(x) = g(x)?

Here are my thoughts on the problem, from beginning to the end:

It seems that $\displaystyle\lim_{n\to\infty}\frac{x}{1+nx^2} = 0$ for all $x$, so $f(x) = 0$.

f_ n'(x) =\frac{1-nx^2}{(nx^2 +1)^2}\;, so \lim_{n\to\infty}f_n'(x) = \lim_{n\to\infty}\frac{\frac1n-x^2}{nx^2+2x^2+\frac1n}= 0 as well.

So far it seems that $g(x) = f(x) = 0$.

When I look at solutions of these sort of problems it seems like the maxima of the sequences are found. By inspections of (f_n'(x)) we see that (f_n'(x)) = 0 when $1-nx^2 = 0$, i.e. when $x = \pm \frac1{\sqrt n}$. From plotting or a few calculations we see that the sequence takes its maxima in the positive solution.

Now, the standard procedure is apparently to insert the maxima into the absolute value $\left|f\left(\frac1{\sqrt n}\right) - f_n\left(\frac1{\sqrt n}\right)\right| = \left|f_n\left(\frac1{\sqrt n}\right)\right| = \frac1{2\sqrt n} \to 0$ when $n$ approaches infinity. From this we see that $f_n$ is uniformly convergent, although my textbook doesn't have any theorems or definitions that incorporate any suprema or maxima (I'd greatly appreciate some insight in this matter).

Similar calculations give that (f_n'(x)) takes its maxima in $\sqrt{\frac3{n}}$, but here |f_n'(x)| = \frac18, so f_n'(x) is not uniformly convergent.

The last question is the one that raised a warning flag. Assuming all my calculations are correct, f'(x) = g(x) for all $x$ since both are the zero function. I think it is suspicious how trivial my answer is.

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    $(f_n ')$ is not uniformly convergent on intervals that don't contain the origin, correct?2012-03-09

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