Nullity and rank of a linear transformation over a function space I have no problem computing nullity and rank of a linear transformation over a finite-dimensional linear space such as $\mathcal{R}^n$.
However, the following exercise from Apostol's Calculus asks for the nullity and rank of a linear transformation over an infinite-dimensional functional space, and I have no idea where to start:

Let $V$ be the linear space of all real functions continuous on
  $[a,b]$. If $f\in V$, $g=T(f)$ means that
$$ g(x) = \int_a^b f(t)\sin(x-t)\,dt \quad \text{for $\;\leq x\leq b$}
> $$

The exercise first asks to prove that the transformation is linear, and this part was easy. I can see intuitively that there are 'all sorts' of functions that will make this integral $0$, so at least I can understand why the nullity is infinite, but I don't know how to prove it. As for the rank, I have absolutely no idea why it's supposed to be $2$. 
 A: I think it helps to observe that
$\sin(x - t) = -\cos x \sin t + \cos t \sin x = \sin x \cos t - \cos x \sin t; \tag 1$
then 
$g(x) = Tf(x) = \displaystyle \int_a^b f(t) \sin(x - t) \; dt = \int_a^b f(t)( \sin x \cos t - \cos x \sin t) \; dt$
$= \displaystyle \sin x \int_a^b f(t) \cos t \;dt - \cos x \int_a^b f(t) \sin t \; dt; \tag 2$
we see that
$0 = g(x) = Tf(x) \tag 3$
precisely when
$\displaystyle \int_a^b f(t) \cos t \;dt = \int_a^b f(t) \sin t \; dt = 0; \tag 4$
since an infinite dimensional subspace of $V$ satisfies (4), the nullity of $T$ is infinite.  Since 
$\dim(\text{span}(\{\cos x, \sin x \}) = 2, \tag 5$
and $\cos x$, $\sin x$ are linearly independent, the rank of $T$ is $2$.
Note: see also this answer.  End of Note.
A: Given linear operator $T:X\rightarrow Y$ where both $X$ and $Y$ are inner product spaces, if there exist two linear independent sets $\{f_{1},...,f_{m}\}\subseteq X$ and $\{g_{1},...,g_{m}\}\subseteq Y$ such that 
\begin{align*}
Tf=\sum_{j=1}^{m}\left<f,f_{j}\right>g_{j},
\end{align*}
then $T$ is of finite rank $m$.
The proof is the following. First the range $R(T)$ of $T$ is clearly contained in the span $L(g_{1},...,g_{m})$ of $\{g_{1},...,g_{m}\}$. Now fix a $j_{0}\in\{1,...,m\}$, there exists an $h_{j_{0}}\in L(f_{1},...,f_{m})$ such that $\left<h_{j_{0}},f_{j_{0}}\right>\ne 0$ and $\left<h_{j_{0}},f_{j}\right>=0$ for all $j\ne j_{0}$. Then $T(h_{j_{0}})=\left<h_{j_{0}},f_{j_{0}}\right>g_{j_{0}}$, so all $g_{j}$ are contained in $R(T)$, this shows that $R(T)=L(g_{1},...,g_{m})$. Since $\{g_{1},...,g_{m}\}$ is linearly independent, so $T$ is of finite rank $m$.
Now plug your example to be $f_{1}(t)=\cos(t)$, $f_{2}(t)=\sin t$, $g_{1}(x)=\sin x$, $g_{2}(x)=-\cos x$ and the inner product to be $\left<u,v\right>=\displaystyle\int_{a}^{b}u(s)v(s)ds$.
Having claim that $T$ is of finite rank $2$, since we have the formula $\dim\ker T+\dim R(T)=\dim X$ (in this case $X$ is the $V$), $\ker T$ is infinite dimensional. Note that the formula is interpreted as set-theoretic cardinals.
A: The fact that $\dim \ker  T = \infty$ follows from the rank nullity theorem.
Pick $n$ linearly independent points $f_1,...,f_n$ and let $V_n = \operatorname{sp} \{ f_k\} \subset V$. Let $T_n = T \mid_{V_n}$
Then apply the rank nullity theorem to $T_n: V_n \to V$ to get
$\dim \ker T_n= \dim V_n - \dim {\cal R} T_n \ge n -2 $.
Since
$\ker T_n \subset \ker T$ we see that $\dim \ker T = \infty$.
For example, we could choose $f_k(t) = t^k$ for $k=0,...,n-1$.
