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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 $f+g$.

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    $\begingroup$ 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, 2022 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, 2022 at 0:39
  • $\begingroup$ @EliasKemperFilho You might be interested in a simple answer to your question, just posted. $\endgroup$ May 7, 2022 at 18:27

2 Answers 2

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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*}

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  • $\begingroup$ Nice answer! I just need understand why the first innequality below "From this and from (∗∗)" is true. $\endgroup$
    – Kempa
    May 3, 2022 at 1:04
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    $\begingroup$ Thanks! Using $h_1(t) = |f(t)|+|g(t)|$ and $h_2(t) = |f(t)+g(t)|$ in $(**)$ $\endgroup$
    – V4k0s
    May 3, 2022 at 1:07
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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.

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