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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.

roman
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  • You can always write $f(x) = \varphi(x) + \frac{a}{T}x$ - the trick is to prove that $\varphi(x)$ is periodic. You say "Given $\varphi(x)$ is periodic,..." but that's what you want to prove. So do not replace $\varphi(x+T)$ by $\varphi(x)$: keep it as is and keep going. – NickD Aug 07 '18 at 21:11
  • @NickD I see your point, however the contradiction raised from $f(x+T) = f(x)$ is very confusing to me. I see how i could express $\varphi (x)$ in terms of $f(x)$, but still not sure why i got $f(x)$ is periodic – roman Aug 07 '18 at 21:16
  • Does this answer your question? functional equation $f\left(x\right)+a=f\left(x+b\right)$ – shangq_tou Feb 28 '21 at 17:19

1 Answers1

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

Tsemo Aristide
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