2
$\begingroup$

Consider the space $C\bigl([a,b];\mathbb{R}\bigr)$ equipped with the $\sup$ norm. Define the operator $$\mathfrak{f} : C\bigl([a,b];\mathbb{R}\bigr) \to C\bigl([a,b];\mathbb{R}\bigr) \: \text{by} \ \ \mathfrak{f}(\varphi)(t) = \int_{a}^{b}(\varphi(s))^{3} ds \cdot \varphi(t), \ \ \text{for} \ \varphi\in C\bigl([a,b];\mathbb{R}\bigr)$$

  • Now for a given $\chi\in C\bigl([a,b];\mathbb{R}\bigr)$,I want to find a linear operator $\mathscr{L} : C\bigl([a,b];\mathbb{R}\bigr) \to C\bigl([a,b];\mathbb{R}\bigr)$ satisfying $$\lim_{||\varphi||_{\infty}\to 0} \: \frac{\mathfrak{f}(\chi+\psi)-\mathfrak{f}(\chi)-\mathscr{L}\varphi}{||\varphi||_{\infty}}=0.$$

  • I also want to show $\mathscr{L}$ is continuous. I know that it suffices to show $\mathscr{L}$ is bounded.

  • Also I want to calculate the derivative of $D\mathfrak{f}(\chi)$ of $\mathfrak{f}$ at $\chi$ $?$

A solution would be of great help.

  • 0
    First, we should denote $(\mathfrak f(\varphi))(t)$, since $\varphi(t)$ is a real number, and $\mathfrak f$ acts on functions, not numbers. I think you mean "$\mathcal L$ is continuous" not $\chi$.2012-12-15
  • 0
    @DavideGiraudo Thanks, is it ok now.2012-12-15
  • 0
    There still is a typo in the second item in the list ($\mathcal L$ instead of $\chi$), and the map $\mathfrak f$ is not linear (so the title needs an edit).2012-12-15
  • 0
    @DavideGiraudo: Thanks.2012-12-15
  • 0
    Expand $(\varphi+h)^3$, and multiplying by $\varphi+h$, we can remove terms in $h^j$, $j\geqslant 2$. What is the linear part in $h$?2012-12-15
  • 0
    @DavideGiraudo: I am sorry I dont get it. why are you doing $(\varphi+h)^{3}$. I guess you are calculating $\mathfrak{f}(\varphi +h)(t)$, but why are you doing that.2012-12-15
  • 0
    I am an Amateur here. Learning on my own. Thats why I asked for a solution. I apologize.2012-12-15
  • 0
    There is a typo in the difference quotient. It should be $\varphi$ instead of $\psi$ (I guess!).2012-12-16

1 Answers 1