Given the Hilbert space $L^2([-1,1])$ endowed with the usual inner product, consider the following operator:
$$Tf:=\int_{-1}^1f(x)e^x \mathrm dx $$
Let $N:=\ker T$, find an explicit formula to find the projection on $N$ and $N^{\perp}$.
My attempt
First of all, I noticed that $T:(L^2([-1,1]),\|\cdot\|_2)\to(\Bbb C, |\cdot|)$ is a continuous operator, hence $N=T^{-1}(\{0\})$ is a closed subset; now I was trying to use the orthogonal decomposition theorem.
My idea was to exploit the fact that we can write a generic function $f$ as the sum of an even and an odd function, i.e. $f=f_1+f_2$ where $f_1$ is odd and $f_2$ is even. I take this route because I wanted to exploit the symmetry of the interval $[-1,1]$; given that the integral of an odd function over a symmetric intervall is $0$, I tried to find the projection on $N$ using odd functions. So I tried for example to consider $e^{-x}f_1$ so that $T(e^{-x}f_1)=0$, but the problem is that I can't go back to the original function $f$.
So my problem boils down to write $f$ as $f=g+h$ where $g\in N$, $h\in N^{\perp}$ and $\langle g,h\rangle=0$.
Any idea or hint is appreciated.
Best Answer
Define the inner product $\langle f,g\rangle = \int_{-1}^{1} f(x)\overline{g(x)}dx$ and note that
$$Tf = \langle f,e^x\rangle.$$
The kernel $N=\ker T = \{ f\in L^2\ | \ \langle f,e^x\rangle = 0 \}$ has codimension 1 and $e^x \in N^{\perp}$.
Every function $f\in L^2$ can be decomposed as
$$f = \left(f-\frac{\langle f,e^x\rangle}{\langle e^x, e^x\rangle}e^x\right)+ \frac{\langle f,e^x\rangle}{\langle e^x,e^x\rangle }e^x.$$
The first term is in $N$ and the second is in $N^{\perp}$. Using the fact that
$$\begin{align}\langle e^x,e^x\rangle &=\int_{-1}^{+1} e^{2x}dx\\ & = \frac{1}{2}\left(e^2-e^{-2}\right)\\ & = \sinh(2)\end{align}$$ the explicit formula for the projection onto the kernel is
$$\text{Proj}_N(f) =f - \left(\dfrac{\int_{-1}^{+1}f(x)e^x dx}{\sinh(2)}\right)e^x$$ and $$\text{Proj}_{N^\perp}(f)= f - \text{Proj}_N(f)$$