Invertibility and inverse of a linear operator

Question

Being an undergraduate student of functional analysis, I am trying to show that the following linear operator is invertible and find its inverse.

Let $C[0,1]$ be the space of real-valued continuous function with the usual supremum norm. Let $A:C[0,1]\to C[0,1]$ be a linear operator defined as: $$Ax(t)=\int_0^1 x(s)ds + x(t).$$ I need to show that $A$ is invertible and then find its inverse. I know that I need to show that $A$ as a map is bijective in order to show invertibility. But, I can't even get my head around showing it an injection.

Answer 1

The linear map $A$ is injective if and only if $\ker A=\{0\}$. But, if $x\in C([0,1])$, $$ x\in\ker A\iff\left(\int_0^1x\right)+x=0. $$ So, $x$ is a constant function (it is equal to $-\int_0^1x$). But, for any real number $k$, $\left(\int_0^1k\right)+k=2k$, and $2k=0\iff k=0$. Therefore, $x$ is the null function.

Answer 2

Once we determine the inverse operator, bijectivity will follow automatically.

Let $(Px)(t)=\int\limits_0^1f(s)\,ds.$ Observe that $P^2=P$ and $A=I+P.$ Hence $$A^{-1}=[(I-P)+2P]^{-1}=I-P+{1\over 2}P=I-{1\over 2}P$$ i.e. $(A^{-1}x)(t)=x(t)-{1\over 2}\int\limits_0^1x(s)\,ds$