Show that $W$ is a subspace of $\mathcal{F}(\mathbb{R},\mathbb{R})$. Let
$\mathcal{F}(\mathbb{R},\mathbb{R})$ the linear space of functions of $\mathbb{R}$ to $\mathbb{R}$ ,
$\displaystyle W=\left\{f\in\mathcal{F}:\int_{-\infty}^\infty\frac{|f(x)|^3}{(1+|f(x)|)^2}\,dx<\infty\right\}$.
Show that $W$ is a subspace of $\mathcal{F}(\mathbb{R},\mathbb{R})$.
My trouble is for $f+g$.
 A: Note that $\mathcal{F} \supset W \neq \emptyset$. Let $f, g \in W$, $\lambda \in \mathbb{R}$ and
$$
F = \int_{-\infty}^{\infty} |f(t)|^3(1+|f(t)|)^{-2} dt < \infty,
$$
$$G = \int_{-\infty}^{\infty} |g(t)|^3(1+|g(t)|)^{-2} dt < \infty.
$$

*

*$f+g \in W$.

Claim If $a,b \in [0,\infty)$ and $a\geq b$, then
\begin{equation}
\frac{a^3}{(1+a)^2}\geq \frac{b^3}{(1+b)^2}. \ \ \ (*)
\end{equation}
Proof: Notice that
$$
\frac{d}{dx}\left(\dfrac{x^3}{(1+x)^2}\right) = \frac{x^2(x+3)}{(x+1)^ 3} \geq 0,\ \ \ \forall x \in [0,\infty).
$$
Hence, $\dfrac{x^3}{(1+x)^2}$ is increasing in $[0,\infty)$ and the desired inequality follows.
Given $h_1, h_2: \mathbb{R}\to [0,\infty) $ with $h_1(t)\geq h_2(t)\ \forall t\in \mathbb{R}$, we have by $(*) $ that
\begin{equation}
\int_{-\infty}^{\infty}\frac{|h_1(t)|^3}{(1+|h_1(t)|)^2}dt \geq \int_{-\infty}^{ \infty}\frac{|h_2(t)|^3}{(1+|h_2(t)|)^2}dt. \ \ \ (**)
\end{equation}
By the triangular inequality
$$
        |f(t)+g(t)| \leq |f(t)|+|g(t)|, \ \ \ \forall t \in \mathbb{R}.
$$ From this and from $(**)$
\begin{align*}
&\int_{-\infty}^{\infty}\frac{|f(t)+g(t)|^3}{(1+|f(t)+g(t)|)^2}dt \\&\leq\int_{-\infty}^{\infty}\frac{(|f(t)|+|g(t)|)^3}{(1+|f(t)|+| g(t)|)^2}dt\\& = \int_{-\infty}^{\infty}\frac{|f(t)|^3+|g(t)|^3+3|f (t)g(t)|(|f(t)|+|g(t)|)}{(1+|f(t)|+|g(t)|)^2}dt\\&= \int_{-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)+|g(t)|)^2}dt \ +\ \int_ {-\infty}^{\infty}\frac{|g(t)|^3}{(1+|f(t)|+|g(t)|)^2}dt\ +\ 3\int_ {-\infty}^{\infty}\frac{|f(t)g(t)|(|f(t)|+|g(t)|)}{(1+|f(t)|+ |g(t)|)^2}dt.
    \end{align*}
Notice that
\begin{equation}
\int_{-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)|+|g(t)|)^2}dt\leq\int_ {-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)|)^2}dt = F < \infty
\end{equation}
and similarly
\begin{equation} \label{desig14} \int_{-\infty}^{\infty}\frac{|g(t)|^3}{(1+|f(t)|+|g(t) |)^2}\leq G < \infty.
\end{equation}
Thus, it is sufficient to prove that
$$
    H= \int_{-\infty}^{\infty}\frac{|f(t)g(t)|(|f(t)|+|g(t)|)}{(1+|f( t)|+|g(t)|)^2}dt<\infty
    $$
to complete the desired result. Define $h:\mathbb{R}\to \mathbb{R}$ by
$$
    h(t)=\max(|f(t)|,|g(t)|)$$
and note that
$$
    |h(t)|^3\leq|f(t)|^3+|g(t)|^3, \ \ \forall t \in \mathbb{R}.
    $$
Also, for every $t$
$$
    |f(t)g(t)|(|f(t)|+|g(t)|)\leq |h(t)h(t)|(|h(t)|+|h(t) |)=2|h(t)|^3.
    $$
Finally
\begin{align*}
 H&\leq\int_{-\infty}^{\infty}\frac{2|h(t)|^3}{(1+|f(t)|+|g(t)|)^2}dt \\&\leq2\int_{-\infty}^{\infty}\frac{|f(t)|^3+|g(t)|^3}{(1+|f(t)|+| g(t)|)^2}dt\\&=2\int_{-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)|+ |g(t)|)^2}dt \ +\ 2\int_{-\infty}^{\infty}\frac{|g(t)|^3}{(1+|f(t)|+ |g(t)|)^2}dt \leq 2F+2G<\infty.
\end{align*}
Therefore
\begin{align*}
\int_{-\infty}^{\infty}\frac{|f(t)+g(t)|^3}{(1+|f(t)+g(t)|)^2}dt <\infty \Rightarrow f+g \in W.
\end{align*}
A: Observe that
\begin{align*}
{1\over 4}|f|^3\le {|f|^3\over (1+|f|)^2}\le |f|^3 & \qquad|f|\le 1\\
{1\over 4}|f|\le  {|f|^3\over (1+|f|)^2}\le |f| & \qquad |f|>1
\end{align*}
Therefore
$${1\over 4}\min(|f|,|f|^3)\le {|f|^3\over (1+|f|)^2}\le \min(|f|,|f|^3)$$
Hence $$f\in W \iff \int \min(|f|,|f|^3)<\infty
$$
Assume $f,g\in W.$ Then
$$\displaylines{\min(|f+g|,|f+g|^3) \le \min(|f|+|g|,(|f|+|g|)^3)\\ \le \begin{cases} 8\min(|f|,|f|^3) & |f|\ge |g|\\
8\min(|g|,|g|^3) & |f|\le |g|
\end{cases}\le 8\min(|f|,|f|^3)+8\min(|g|,|g|^3)]}$$
Therefore $f+g\in W.$
Remark The method can capture many other similarly defined subspaces.
