$\mathcal{F}(\mathbb{R},\mathbb{R})$ 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 $f+g$.

  • 1
    $\begingroup$ 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? $\endgroup$
    – 1Rock
    May 3 at 0:04
  • $\begingroup$ 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. $\endgroup$
    – Kempa
    May 3 at 0:39
  • $\begingroup$ @EliasKemperFilho You might be interested in a simple answer to your question, just posted. $\endgroup$ May 7 at 18:27

2 Answers 2


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 \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*}

  • $\begingroup$ Nice answer! I just need understand why the first innequality below "From this and from (∗∗)" is true. $\endgroup$
    – Kempa
    May 3 at 1:04
  • 2
    $\begingroup$ Thanks! Using $h_1(t) = |f(t)|+|g(t)|$ and $h_2(t) = |f(t)+g(t)|$ in $(**)$ $\endgroup$
    – Vakos
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.