Determine if the following sequence converge in the quadratic mean

Question

"For integer $n$ let $f_n(x) = \dfrac{1}{\sqrt{1 + nx^2}}$

say if the sequence $f_n$ converges in quadratic mean."

This is what I have concluded so far:

$$\lim\limits_{n\rightarrow \infty}f_{n}(0) = 1$$

For $x \neq 0$

$$\lim\limits_{n\rightarrow \infty}f_{n}(x) = 0$$

For convergence in quadratic mean we must show that

$$ \lim\limits_{n\to\infty}||f_{n}-f||_{2}$$

where $f =\lim\limits_{n\to\infty}f_n.$ and $||\cdot||_{2}$ is the $L^{2}$ norm.

For $x=0$ it is obvious. What about when $x\neq 0$?

Answer 1

The sequence $(f_n)$ is pointwise convergent to the function $f$ defined on the interval $[0,+\infty)$ by $$f(x)=\left\{\begin{array}\\0&\text{if}\ x>0\\ 1&\text{if}\ x=0\end{array}\right.$$ which's equal to the zero function almost everywhere and since we have $$|f_n(x)|^2=\frac{1}{1+nx^2}\leq \frac{1}{1+x^2}=g(x)\quad\forall n\geq 1$$ and the function $g$ is integrable on the interval $[0,+\infty)$ then by the dominated convergence theorem the sequence $(f_n)$ converges to $f$ in quadratic mean.

Answer 2

The result is a direct consequence of Dominated Convergence theorem.