# 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 $$f+g$$.

• Are you having trouble showing that $\lambda f \in W$ for $f \in W$ (try cases for $|\lambda|\ge 1$ and $|\lambda|<1$), or that $f + g \in W$ for $f,g \in W$? Can you post your working so far, so we can suggest a direction? May 3 at 0:04
• Indeed is for $f+g$ my difficulty. I'm trying yet, but is sketch. I know that need use innequations, then I'll see what to do. May 3 at 0:39
• @EliasKemperFilho You might be interested in a simple answer to your question, just posted. May 7 at 18:27

Note that $$\mathcal{F} \supset W \neq \emptyset$$. Let $$f, g \in W$$, $$\lambda \in \mathbb{R}$$ and $$F = \int_{-\infty}^{\infty} |f(t)|^3(1+|f(t)|)^{-2} dt < \infty,$$ $$G = \int_{-\infty}^{\infty} |g(t)|^3(1+|g(t)|)^{-2} dt < \infty.$$

1. $$f+g \in W$$.

Claim If $$a,b \in [0,\infty)$$ and $$a\geq b$$, then $$$$\frac{a^3}{(1+a)^2}\geq \frac{b^3}{(1+b)^2}. \ \ \ (*)$$$$ Proof: Notice that $$\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 $$$$\int_{-\infty}^{\infty}\frac{|h_1(t)|^3}{(1+|h_1(t)|)^2}dt \geq \int_{-\infty}^{ \infty}\frac{|h_2(t)|^3}{(1+|h_2(t)|)^2}dt. \ \ \ (**)$$$$ By the triangular inequality $$|f(t)+g(t)| \leq |f(t)|+|g(t)|, \ \ \ \forall t \in \mathbb{R}.$$ From this and from $$(**)$$ \begin{align*} &\int_{-\infty}^{\infty}\frac{|f(t)+g(t)|^3}{(1+|f(t)+g(t)|)^2}dt \\&\leq\int_{-\infty}^{\infty}\frac{(|f(t)|+|g(t)|)^3}{(1+|f(t)|+| g(t)|)^2}dt\\& = \int_{-\infty}^{\infty}\frac{|f(t)|^3+|g(t)|^3+3|f (t)g(t)|(|f(t)|+|g(t)|)}{(1+|f(t)|+|g(t)|)^2}dt\\&= \int_{-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)+|g(t)|)^2}dt \ +\ \int_ {-\infty}^{\infty}\frac{|g(t)|^3}{(1+|f(t)|+|g(t)|)^2}dt\ +\ 3\int_ {-\infty}^{\infty}\frac{|f(t)g(t)|(|f(t)|+|g(t)|)}{(1+|f(t)|+ |g(t)|)^2}dt. \end{align*} Notice that $$$$\int_{-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)|+|g(t)|)^2}dt\leq\int_ {-\infty}^{\infty}\frac{|f(t)|^3}{(1+|f(t)|)^2}dt = F < \infty$$$$ and similarly $$$$\label{desig14} \int_{-\infty}^{\infty}\frac{|g(t)|^3}{(1+|f(t)|+|g(t) |)^2}\leq G < \infty.$$$$ Thus, it is sufficient to prove that $$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*}

• Nice answer! I just need understand why the first innequality below "From this and from (∗∗)" is true. May 3 at 1:04
• Thanks! Using $h_1(t) = |f(t)|+|g(t)|$ and $h_2(t) = |f(t)+g(t)|$ in $(**)$ May 3 at 1:07

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.