Given $f(x+T) = f(x) + a$, prove that $f(x) = \varphi(x) + {a \over T}x$, where $\varphi(x)$ is periodic with period $T$

Question

For some function $f(x)$ there exist numbers $T \ne 0$ and $a$ such that for any $x \in D(f)$:

$$ x+T \in D(f) \\ x-T \in D(f) $$

and $f(x+T) = f(x) + a$.

$D(f)$ denotes the domain of $f$.

Prove that $f(x) = \varphi (x) + {a \over T}x$, where $\varphi (x)$ is a periodic function with period $T$.

Here is my try but I have strong doubts about its validity.

Assume that $f(x) = \varphi (x) + {a \over T}x$. Lets see how $f(x)$ behaves in $x+T$:

$$ f(x+T) = \varphi (x+T) + {a \over T}(x+T) $$

Given $\varphi (x)$ is periodic, then $\varphi (x+T) = \varphi (x)$. Therefore:

$$ \begin{align} f(x+T) &= \varphi (x) + {a \over T}(x+T) \\ &=\varphi (x) + {ax \over T} +a \end{align} $$

But $f(x+T) = f(x) + a$ and then:

$$ f(x+T) = f(x) + a \\ f(x) + a =\varphi (x) + {ax \over T} + a \iff \\ \iff f(x) = \varphi (x) + {ax \over T} + a - a $$

So finally I arrived at where I started. Moreover the above shows that $f(x)$ is periodic which contradicts the initial statement that $f(x+T) = f(x) + a$

One may notice that $f(x + nT) = f(x) + na$ but I don't see how I could use that fact. What is the correct way of proving the above?

Please note that the restrictions in the question section is everything i'm given. So the function is just known to have some domain $D(f)$ and both $T$ and $a$ are not restricted to any particular set of numbers. Also that is from pre-calc section.

Answer 1

Let $\phi(x)=f(x)-{a\over T}x$, $\phi(x+T)=f(x+T)-{a\over T}(x+T)=f(x)+a-{a\over T}(x+T)=f(x)-{a\over T}x=\phi(x)$.