Proving a Sequence's Uniform Convergence

Question

Another homework problem that's been giving me headaches for about a week now.

Prove that the following sequence of functions $(f_n)$ converges uniformly on the interval $[1,2]$:

$$f_n(x) = \frac {nx^2 - 2}{x^4 + nx}.$$

Then, find the following integral on the same interval and justify your answer:

$$ \lim_{n\to\infty} \int_1^2 \frac {nx^2 - 2}{x^4 + nx}\mathrm dx.$$

Any help on this would be appreciated.

Answer 1

$f_n(x) = \frac {nx^2 - 2}{x^4 + nx}$. So make a guess that $f_n \to x$. Then try to prove that for all $\epsilon \gt 0$, there exists natural $N$ such that $n \gt N, x \in [1,2] \implies |f_n(x) - x | \lt \epsilon$.

Well $\frac{nx^2 - 2}{x^4 + nx} - x = \frac{nx^2 - 2 - x^5-n x^2 }{x^4 + nx} = $ top and bottom of $x$ term multiplied by $x^4 + nx. \ $ Then $ = -\frac{x^5 + 2}{x^4 + nx}$. Clearly this can be made arbitrarily small, but let's prove it. Max of $|x^5 + 2|$ on $[1,2]$ is $M_1 = 2^5 + 2$. So $|f_n(x) - x | = |\frac{x^5 + 2}{x^4 + nx}| \leq \frac{M_1}{x^4 + nx}$ Since $[1,2]$ is an all positive interval we have that $nx \lt x^4 + nx$, so we have $RHS \lt \frac{M_1}{nx}$. And for any $n$, the minimum of the denominator is $1\cdot n$, so we have $\frac{M_1}{nx} \lt \frac{M_1}{n}$. Choose $N$ small enough so that $M_1/N \lt \epsilon$.

Answer 2

Hints:

1) $$ f_n(x) = x\left( \frac{ 1 - \frac{2}{nx^2}}{1 + \frac{x^3}{n}}\right)$$

2) Since $f_n \to f$ uniformly on $[a,b]$ then $$ \int_a^b |f_n(x) - f(x)|dx \leq (b-a)\cdot \sup_{x \in [a,b]} \{|f_n(x) - f(x)|\}$$ might be a useful inequality.

Answer 3

1. We can identify the pointwise limit: it's $f(x)=x$

2. $f_n(x)-f(x)=\frac{nx^2-2-x^5+nx^2}{x^4+nx}=-\frac{2+x^5}{x^4+nx}$. So if $x\in [1,2]$, $|f_n(x)-f(x)|\leqslant \frac{2+2^5}{n}$.

3. Justify we can put the limit under the integral.