Kernel and Image are orthogonal ==> They are supplementary? Let $H$ be a Hilbert space and $f:H\rightarrow H$ an endomorphism. Suppose the image and kernel of $f$ are orthogonal. Can I conclude that they are supplementary? (Namely $H = \mathop{Ker}(f) \oplus \mathop{Im}(f)$)
If $H$ is finite dimensionnal, this follows from the rank theorem. If it is not however, I am not sure how to conclude. Am I missing an assumption?
I'm not sure this is relevant, but in my specific case, $f$ has the form $f = id - r$, where $r$ is and isometry. Does the result hold with this added assumption?
 A: They are not, even if $f=id-r$ where $r$ is a surjective isometry.
Counterxample: $H=\ell^2(\mathbb Z)$, $r(u)_n:=u_{n+1}$.
Then, $\ker f=\{0\}$ (the only constant sequence in $\ell^2$ is $0$) but we shall prove that $f$ is not onto:
let $v_n=\frac1n$ if $n>0$, and $v_n=0$ if $n\le0$. Let us prove by contradiction that $v\notin\mathrm{im}f$.
If $v=f(u)$ then $\forall n>0\quad u_n=u_0-\sum_{0<k<n}\frac1k$, which contradicts $u\in\ell^2.$
A: The answer is negative, even in your special case. Consider $H := \ell^2(\mathbb{N})$ and $r: H \to H$ the right-shift operator, i.e.,
$$r(x)_1 = 0, \quad r(x)_j = x_{j-1}, \quad j = 2,\ldots.$$
Clearly $r$ is an isometry and it is well-known that the spectrum of $r$ is the unit disk. Now I claim that $f = \mathrm{id}-r$ is injective. Indeed, if $f(x) = 0$, then
$$0 = f(x)_1 = x_1 - r(x)_1 = x_1.$$
Moreover,
$$0 = f(x)_j = x_j - r(x)_j = x_j - x_{j-1}$$
for $j \geq 2$. This shows $x = 0$ by induction. So $f$ is injective and thus, trivially, $\mathrm{Ker}(f)$ is orthogonal to $\mathrm{Im}(f)$. On the other hand $\mathrm{Ker}(f) \oplus \mathrm{Im}(f)$ cannot be $H$ as $f$ is not invertible ($1$ is in the spectrum of $r$).
A: For a counterexample, take the Hilbert space $H=\ell^2(\mathbb R)$ of square summable sequences endowed with the inner product
$$\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i
$$ and for $f$ the left shift operator: $$f((x_1, x_2, \dots)) = (x_2, x_3, \dots)$$
