Finding the rank and nullity of $T$ with $T(f)(x) = \int_a^b f(t) \sin(x-t) \,\mathrm{d}t$

Question

Problem

Let $V$ be the linear space of all the real functions continuous on $[a,b]$. If $f\in V$, $g=T(f)$ means that $\int_{a}^{b} f(t) \sin(x-t) \,\mathrm{d}t$ for $a\leq x \leq b $. Describe its null space and range, and compute its nullity and rank.

Attempt

$\text{range} = \{ g(x) = \int_{a}^{b}f(t) \sin(x-t) \,\mathrm{d}t, a\leq x \leq b \}$

$\text{null space} = \{ f(x) \mid \int_{a}^{b} f(t) \sin(x-t) \,\mathrm{d}t = 0 , a\leq x \leq b \}$

$\text{nullity} = \infty$

Doubt

How to calculate the rank? I have the inkling that the rank should be zero since a function $f$ is transformed into a constant function, but I am not able to prove it.

Also, is the nullity theorem only valid for finite dimensional vector spaces or there is there something equivalent for infinite dimensional vector spaces?

Null space is edited after suggestion

Answer 1

Your range, while technically correct, is probably not the answer they're looking for. Your null space is wrong.

Hint: $T(f)$ is a function of $x$; which functions can we get?

Hint: The domain of $T$ is the space of all continuous real functions on $[a,b]$, and the null space is a subset of this domain. Which functions $f$ satisfy $T(f) = 0$?

You shouldn't be using the rank-nullity theorem. Instead, once you have a suitable description of the null space and range, it should be easy to compute the dimension of the null space (the nullity) and the dimension of the range (the rank).

Answer 2

The trick is to rewrite the integral as $$ Tf = \int_a^b f(t)\sin(x-t)dt \\ = \int_a^b f(t)\cos(t)dt \sin(x)-\int_{a}^{b}f(t)\sin(t)dt\cos(x). $$ So the range of $T$ has basis $\{\sin(x),\cos(s)\}$. The null space of $T$ consists of all $f$ such that $\int_a^b f(t)\sin(t)dt=0=\int_a^b f(t)\cos(t)dt$.