I am taking a first course in functional analysis and I am struggling with working with operator norms.
Specifically, consider $L^2(\mathbb{R})$ with the linear operator $\Lambda:L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ defined by $(\Lambda f)(x)=f(|x|)$.
I have the following equality $$\|\Lambda f\|_{L^2(\mathbb R)}\doteq \Big[\int_{-\infty}^{\infty}|f(|x|)|^2dx\Big]^{\frac{1}{2}}=\Big[\int_{-\infty}^{0}|f(-x)|^2dx+\int_{0}^{\infty}|f(x)|^2dx\Big]^{\frac{1}{2}}=\Big[2\cdot\int_{0}^{\infty}|f(x)|^2dx\Big]^{\frac{1}{2}}$$
I am struggling with providing a rigorous/nice way to compute the kernel and range of $\Lambda$.
From the above calculation I just convinced myself that $\int_{0}^{\infty}|f(x)|^2dx=0$ which is equivalent to $\text{ker}(\Lambda)=\{f\in L^2(\mathbb R):f=0, a.e. x\geq 0\}$. Is there a way to actually compute this explicitly?
For the range, I had an even tougher time. I believe it has to do with how we define $f(|x|)$ which equals $f(x)$ on $x\geq 0$ and $f(-x)$ on $x<0$. Our professor gave us a solution that says that $\text{range}(\Lambda)=\{f\in L^2(\mathbb R): f(x)=f(-x), a.e. x\in \mathbb R\}$.
Thank you for your help.
If $g=\Lambda(f)$ then $g(-x)=f(|-x|)=f(|x|)=g(x)$ almost everywhere. Conversely, if $g(-x)=g(x)$ almost everywhere define $f=g$ so $f(|x|)=g(|x|)=g(x)$. Hence the range consists precisely of those functions $g \in l^{2}$ such that $g(-x)=g(x)$ almost everywhere. I think you have already found the kernel correctly. Let me assure you that you are in the right track and kernel of $\Lambda$ consists of those $f \in l^{2}$ which are such that $f(x)=0$ almost everywhere 0n $(0,\infty)$.