For $f \in L^2(\mathbb{R})$ and $b \in \mathbb{R}$, define a modulation operator $E_b$ from $L^2(\mathbb{R})$ to itself as:
$E_b f(x) = e^{2\pi i b x}f(x)$ .
Then the question is: for $a \in \mathbb{R}$, what is $E_bf(x-a)$?
I don't know if this is standard notation or not because I'm a beginner, but this is how the operator is defined in a book I am reading. However, it seems to me that the definition allows this question to be answered in two ways.
First, let $y = x-a$ . Then
$E_bf(x-a) = E_bf(y) = e^{2\pi i b y}f(y) = e^{2\pi i b (x-a)}f(x-a)$ .
Second, let $y(x) = f(x-a)$ . Then
$E_bf(x-a) = E_by(x) = e^{2\pi i b x}y(x) = e^{2\pi i b x}f(x-a)$ .
The two answers are not the same. Is there a problem with the notation, or am I doing something wrong?
copper.hat is correct, but let me explain further. The two answers you provided could be written as $$ (E_b f)(x-a) \tag 1 $$ $$ E_b(f(x-a)) \tag 2 $$ However if you think about it, (2) makes no sense. This is because $E_b$ is a function on $L^2(\mathbb R)$, whereas formula (2) is trying to apply $E_b$ to the scalar $f(x-a)$.
Some correct ways to write what you intended with your second answer could be $$ (E_b(x\mapsto f(x-a)))(x) $$ or perhaps better (to avoid the letter $x$ being used in two different contexts) $$ (E_b(z\mapsto f(z-a)))(x) $$ or a shorthand that is commonly used: $$ (E_b(f(\cdot-a)))(x) $$ Anyway these are all rather messy. Also I added an unnecessary set of brackets in these three expressions - for example, the last formula could be written as $$ E_b(f(\cdot-a))(x) $$ because the other ordering of operations $E_b((f(\cdot-a))(x))$ again makes no sense because then $E_b$ is being applied to a scalar.