Given an operator $M$: $L^2([0,1])\rightarrow L ^2([0,1])$: $$M(f)(x) = x^2f(x) $$
I am trying to show if $(I + M)$ is invertible and what $|| (I+M)^{-1} ||$ is. I am aware of the theorem which says if $||M|| < 1$ then $I-M$ and hence $I+M$ is invertible and allows computation of $|| (I+M)^{-1} ||$. But here since $||M|| = 1$, I am stuck about how to proceed. Any help is greatly appreciated.
Let $P=I+M$ and $Qf=\dfrac{1}{1+x^{2}}\cdot f(x)$, it is routine to check that $PQf=f$ and $QP f=f$, so $P$ is algebraic invertible.
We also note that \begin{align*} \|Qf\|_{L^{2}}^{2}=\int_{0}^{1}\dfrac{1}{(1+x^{2})^{2}}|f(x)|^{2}\leq\int_{0}^{1}|f(x)|^{2}dx=\|f\|_{L^{2}}^{2}, \end{align*} so $\|Q\|\leq 1$.
Now we let $f(x)=\chi_{[0,1/n]}(x)$, then $\|f\|_{L^{2}}^{2}=\dfrac{1}{n}$ and that \begin{align*} \|Qf\|_{L^{2}}^{2}=\int_{0}^{1/n}\dfrac{1}{(1+x^{2})^{2}}dx\geq\dfrac{1}{(1+(1/n)^{2})^{2}}\cdot\dfrac{1}{n}=\dfrac{1}{(1+(1/n)^{2})^{2}}\cdot\|f\|_{L^{2}}^{2}, \end{align*} so \begin{align*} \|Q\|\geq\dfrac{1}{1+(1/n)^{2}}, \end{align*} taking $n\rightarrow\infty$, we get $\|Q\|\geq 1$, we conclude that $\|Q\|=1$.