Finding a dominating function

Question

I have been asked to find a dominating function for the following sequence of functions: $$f_{n}(x)=\frac{x n^{3/2}}{1+n^{2}x^{2}}, x\in[0,1]$$ in order to use the dominated convergence theorem.

I have tried but I cannot find a dominating function (I get that $f_{n}(x)\leq \frac{1}{x}$, but this one is not Lebesgue integrable)

Answer 1

The function $g(x) = \frac{x^{3/2}}{1+x^2}$ has a maximum for $x=\sqrt{3}$ with maximum value $\frac{3^{3/4}}{4} < 1$ so

$$f_n(x) = \frac{1}{\sqrt{x}} g(nx) < \frac{1}{\sqrt{x}}$$

and $\frac{1}{\sqrt{x}}$ is Lebesgue integrable on $(0,1]$.

Answer 2

Let $f_n(x)$ be the sequence $f_n(x)=\frac{n^{3/2}x}{1+n^2x^2}$. Then, the derivative $\frac{df_n(x)}{dn}$ is given by

$$\frac{df_n(x)}{dn}=\frac{n^{1/2}x}{2(1+n^2x^2)^2} (3-n^2x^2)\tag 1$$

From $(1)$, we see that $\frac{df_n(x)}{dn}=0$ for $nx=\sqrt{3}$, $\frac{df_n(x)}{dn} >0$ for $nx<\sqrt{3}$, and $\frac{df_n(x)}{dn} <0$ for $nx>\sqrt{3}$. Therefore, $f_n(x)$ has a maximum value at $nx=\sqrt{3}$.

Now, at $nx=\sqrt{3}$, $f_n(x)=\frac{3^{3/4}}{4\sqrt{x}}$. Therefore, the sequence $f_n(x)$ satisfies the inequality

$$f_n(x)\le \frac{3^{3/4}}{4\sqrt{x}} $$

Since we have

$$\int_0^1 \frac{3^{3/4}}{4\sqrt{x}}\,dx=\frac{3^{3/4}}{2}<\infty$$

then by the dominated convergence theorem

$$\lim_{n\to \infty}\int_0^1 f_n(x)\,dx=\int_0^1 \lim_{n\to \infty} f_n(x)\,dx=0$$