Given a function $f(x)$ with $x\in \mathbb{R}$, does the condition
$$f(x+1)=0\text{ for all }x$$
necessarily imply $f(x)\equiv 0$?
I am asking, since e.g. in an instance where $f(x)$ is defined with local support, for example
$$f(x)=\lim_{\epsilon\to 0}\frac{1}{2\epsilon}e^{-\frac{|x|}{\epsilon}}$$
$f(x)$ is only non-zero at the original value $x$ and shifting the argument to $x+1$ takes it away from the local support and makes it vanish without making it zero at the original $x$.
Of course one could say that one could consider a situation instead where we start with $f(x-1)$ and shift the argument to be $x$ and after the shift the expression must vanish just the same, but somehow it seems to me that this only redefines the origin and makes no statement about the value of the function at the original point ($x-1$ in this case).
Does this make any sense, or am I getting confused here?
Yes of course indeed for any $x$ exists $y=x-1$ and then
$$f(x)=f(y+1)=0$$