Computing the $L^2$ norm of a linear operator over a Hilbert space

Question

I am taking a first course in functional analysis and I am struggling with computing the norm of an operator.

Specifically, consider $L^2(\mathbb{R})$ with the linear operator $\Lambda:L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ defined by $(\Lambda f)(x)=f(|x|)$.

$$\|\Lambda f\|_{L^2(\mathbb R)}\doteq \Big[\int_{-\infty}^{\infty}|f(|x|)|^2dx\Big]^{\frac{1}{2}}=\Big[\int_{-\infty}^{0}|f(-x)|^2dx+\int_{0}^{\infty}|f(x)|^2dx\Big]^{\frac{1}{2}}=\Big[2\cdot\int_{0}^{\infty}|f(x)|^2dx\Big]^{\frac{1}{2}}\leq\sqrt{2}\cdot\|f\|_{L^2}$$ I have reached this point and I am tempted to say that $\|\Lambda\|_{L^2}=\sqrt{2}$ but cannot see why. Our professor provided us with a solution that says "equality holds whenever $f$ is supported on the positive half axis." Could someone help me understand what he means by that?

Thank you for your help.

Answer 1

OP has shown

$$\|\Lambda f\|_{L^2(\mathbb R)} = \Big[2\cdot\int_{0}^{\infty}|f(x)|^2dx\Big]^{1/2}\leq\sqrt{2}\cdot\|f\|_{L^2}.$$

Equality holds

\begin{align} \Big[2\cdot\int_{0}^{\infty}|f(x)|^2dx\Big]^{1/2} &= \sqrt{2}\cdot\|f\|_{L^2} \\ \iff \sqrt{\int_{0}^{\infty}|f(x)|^2dx} &= \sqrt{\int_{-\infty}^{\infty}|f(x)|^2dx} = \sqrt{\int_{-\infty}^{0}|f(x)|^2dx + \int_{0}^{\infty} |f(x)|^2dx} \\ \iff \int_{-\infty}^{0}|f(x)|^2dx &= 0 \\ \iff f &= 0 \text{ a.e. on } (-\infty,0] \end{align}

Answer 2

What your prof said is that if $f=f\,1_{[0,\infty)}$, then \begin{align} \|\Lambda f\|_2^2&=\int_{-\infty}^0f(-x)^2\,dx+\int_0^\infty f(x)^2\,dx\\ \ \\ &=\int_0^\infty f(x)^2\,dx+\int_0^\infty f(x)^2\,dx\\ \ \\ &=2\int_0^\infty f(x)^2\,dx\\ \ \\ &=2\|f\|^2_2. \end{align} So, for any such $f$, $$\|\Lambda f\|_2=\sqrt2\,\|f\|_2.$$