# Prove the dominated convergence for $f_n(x)=\frac{x^n}{x^2+3x+2}$

Suppose that we have $f_n:[0,1]\rightarrow\mathbb{R},f_n(x)=\frac{x^n}{x^2+3x+2}$.

• $f_n$ don't have uniform convergence because the pointwise isn't continuous: $$\lim_{n\to\infty}f_n(x)\neq\lim_{n\to\infty}f_n(1)$$

• To prove that $\{f_n\}$ has dominated convergence we need an integrable function $g$ such that $|f_n|\leq g$ and $f_n\rightarrow f$ pointwise as $n\rightarrow\infty$.

But if the pointwise isn't continuous for every $x\in[0,1]$, how we can show the dominated convergence? Or for dominated convergence isn't necessary to have a continuous pointwise limit for every $x\in[0,1]$ ?

• By the way, how can I prove that $f_n\leq g$ where $g$ is integrable ? I suppose $g$ can be $1$ which is integrable on the interval but how can I prove inequality such that $f_n\leq 1$ ? or something else, not necessarily $1$...
• If $x\in[0,1]$, $$f_n(x)=\frac{x^n}{(x+1)(x+2)}\leq \frac{1}{1\cdot 2}.$$ – Jack D'Aurizio Jun 1 '15 at 12:34
• Moreover, since $f_{n+1}(x)\leq f_{n}(x)$, the most natural candidate for a dominating function is just $f_1(x)$. – Jack D'Aurizio Jun 1 '15 at 12:36
• hi Jack, can you give me some explanations for yellow box ? as I said the pointwise convergence isn't continous, but in dominated convergence isn't neccessary this condition ? – Lucas Jun 1 '15 at 12:57

Note that \begin{align*} \frac{x^{n}}{x^{2}+3x+2}\leq \frac{1}{0+0+2}=\frac{1}{2} \end{align*} for all $n\in\mathbb{N}$, and the constant function $\frac{1}{2}$ is integrable on $[0,1]$.
• Thomas, so we can say that $f_n$ is pointwise convergent to $0$, $\forall x\in[0,1]$ or $x\in[0,1)$ ? I mean in dominated convergence not uniform – Lucas Jun 1 '15 at 13:04
• @lucas: The point wise convergence to zero happens for all $x\in [0,1)$. At $x=1$ the limit is not zero, but the single point $x=1$ has measure zero and thus does not contribute to the value of the integral. – T. Eskin Jun 1 '15 at 13:15
• Thomas for $x=1\Rightarrow\lim_{n\to\infty}f_n(x)$ will be $\frac{1}{6}$ I don't understand what you want to say by "$x=1$ has measure zero.." – Lucas Jun 1 '15 at 13:24