Find inverse of an operator and its norm I have an operator $A\in {B(L^2[0,1])}$ and $(Ax)(t) = x(t) - \frac{1}{2}\int_{0}^{1}x(s)ds$ where
$L_2[0,1]$ is a space of functions for which the 2th power of their absolute value is Lebesgue integrable
$B(L^2[0,1])$ - Banach algebra
I need to find norm of my inverse operator in order to find distance to the set of non-invertible operators.
My attempt:

*

*I found $A^{-1}$. $A: x(t) \mapsto x(t)-1/2\int_{0}^{1}x(s)ds = y(t)$ Hence, $(A^{-1}y)(t) = y(t) - e^{t}\int_{0}^{1}e^sy(s)ds$.     $\left\| (A^{-1}y)(t)\right\| = (\int_{0}^{1}\left|y(t)-e^{t}\int_{0}^{1}e^sy(s)ds \right|^2dt)^{1/2}$. But I understand that in $L_2[0,1]$ space it is hard to solve it. It will be easier if our space was continuous functions.


*So, my second idea is to use Fourier series to transform my function: $x(t)-1/2\int_{0}^{1}x(s)ds$. So it will be easier. Or maybe not
Pls, can someone help? Any help appreciated.
 A: The $A^{-1}$ you calculated is wrong. In fact, the inverse operator is given by
$$(A^{-1}y)(t)=y(t)+\int_0^1 y(s)\,ds$$
To see this, denote the integral $\int_0^1 f(t)\,dt$ by $\overline{f}$. Then
$$Ax=x-\frac{1}{2}\overline{x}$$
If $y=Ax$, then
$$\overline{y}=\overline{Ax}=\overline{x}-\frac{1}{2}\overline{x}=\frac{1}{2}\overline{x}$$
hence if $y=Ax$ then
$$y=x-\overline{y}$$
and so
$$x=y+\overline{y}$$
so the inverse operator is given by
$$(A^{-1}y)(t)=y(t)+\int_0^1y(s)\,ds$$
The calculation of the norm of the inverse is then straight-forward. First we note that by the triangle inequality
$$\|A^{-1}y\|=\|y+\overline{y}\|\leq \|y\|+\|\overline{y}\|=\|y\|+|\overline{y}|\qquad (1)$$
and by the Cauchy-Schwartz inequality,
$$|\overline{y}|=\left |\int_0^1y(s)\,ds\right|\leq \left(\int_0^1 |y(s)|^2\,ds\right)^{1/2}=\|y\|$$
Hence, by (1)
$$\|A^{-1}y\|\leq 2\|y\|$$
which implies that
$$\|A^{-1}\|\leq 2$$
Take the constant function $y_0\equiv 1$, which has norm $1$, and
$$\|A^{-1}y_0\|=2=2\|y_0\|$$
which proves that in fact,
$$\|A^{-1}\|=2$$
A: The operator is of the form $$A=I-{1\over 2}P=(I-P)+{1\over 2}P$$ where $P$ is a nontrivial orthogonal projection, in OP case onto constant functions. Therefore $$A^{-1}= (I-P)+2P=I+P$$ In general if $$A=I -\alpha P=(I-P)+(1-\alpha)P $$ for $\alpha \neq 1,$ then $$A^{-1}=(I-P) +{1 \over 1-\alpha } P\quad (*)$$
For $v$ by applying the Pythagorean theorem twice we get $$\|A^{-1}v\|^2=\|(I-P)v\|^2+{1\over |1-\alpha|^2}\|Pv\|^2\\ 
\le {1\over \min\{1,|1-\alpha |^2\}}[\|(I-P)v\|^2+\|Pv\|^2]\\ =
{1\over \min\{1,|1-\alpha |^2\}}\,\|v\|^2 $$ Therefore $$\|A^{-1}\|\le {1\over \min\{1,|1-\alpha |\}}$$
By $(*)$ the norm is attained either on the range of $P$ or on the range of $I-P, $ depending which number $1$ or $|1-\alpha |$ is smaller.
