Dimension of a range space of a linear operator Define $$T: C[0,1] \rightarrow C[0,1]$$ as
$\displaystyle T(f(x))= \int_{0}^{1} \sin(x+y)f(y)dy$, then the dimension of the range space of $T$ is?. Can somebody help me to understand how to approach the problem. If it would have been a finite dimensional space i would have found out the nullity and then substracted it from the dimension of the domain but here this is not the case
 A: Hint: 
\begin{align}
\int^1_0 \sin(x+y)y^n\ dy =&\ \int^1_0 \sin(x)\cos(y) y^n+\cos(x)\sin(y)y^n\ dy \\
=&\ \left(\int^1_0y^n\cos y\ dy\right)\sin x+\left(\int^1_0 y^n \sin y\ dy \right) \cos(x)
\end{align}
So...the dimension of the range is finite.
A: You have
$$
\sin(x+y)=\sin x\,\cos y+ \cos x\,\sin y. 
$$
So 
$$
Tf(x)=\left(\int_0^1\cos y\,f(y)\,dy\right)\,\sin x+\left(\int_0^1\sin y\,f(y)\,dy\right)\,\cos x.
$$
Thus $Tf$ is a linear combination of the two functions $\sin x$ and $\cos x$, and its range has dimension $2$. 
A: Since
$\sin (x + y) = (\sin x) (\cos y) + (\cos x) (\sin y), \tag 1$
we have, for any $f(x) \in C[0, 1]$, 
$T(f(x)) = \displaystyle \int_0^1 (\sin (x + y))f(y)\; dy = \int_0^1 (\sin x) (\cos y)f(y)\;dy + \int_0^1 (\cos x) (\sin y)f(y)\;dy$
$= \displaystyle \sin x \int_0^1  (\cos y)f(y)\;dy + \cos x \int_0^1 (\sin y)f(y)\;dy, \tag 2$
which shows that
$\dim \text{Range}(T) \le 2; \tag 3$
since $\cos x$ and $\sin x$ are linearly independent, it is easy to see that equality must hold in (3), since there exist $g, h \in C[0, 1]$ with
$\displaystyle \int_0^1 (\cos y) g(y) \ne 0, \; \int_0^1 (\sin y) g(y) = 0, \tag{4}$
and
$\displaystyle \int_0^1 (\cos y) h(y) = 0, \; \int_0^1 (\sin y) h(y) \ne 0. \tag{5}$
