Does there exist an analytic everywhere, piecewise-defined function $f$ such that:
$f(x) = g(x)$ for $x < k$
$f(x) = h(x)$ for $x>k$
$f(x) = r$ for $x=k$
With $g \ne h $ ($g$ not the same function as $h$)
If it exists, what is such an example?
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If $g$ and $h$ are not necessarily analytic, then just take an arbitrary analytic function $f$ with $f(k) = r$. Define $g$ and $h$ to be
$$ g(x) := f(x) \chi(k-x) \qquad h(x) := f(x) \chi(x-k) $$
where $\chi$ is a cut-off function, that is: $\chi(x) = 1$ for $x \geq 0$, $\chi(x) = 0$ for $x \leq -1$ and $0 < \chi(x) < 1$ for $-1 < x < 0$. Then you have all your desired properties. You can choose $\chi$ to be continuous, or even smooth.
If $g$ and $h$ are analytic, then the answer is no. Real analytic functions have the unique continuation property. That is, if a real analytic function is zero on an open set, it must be zero everywhere. Then since $f$ is everywhere analytic, and $g$ is everywhere analytic, their difference $f-g$ is also analytic, and vanishes for $x < k$. This implies that $f=g$ everywhere. Similarly $f =h$ everywhere and $g$ and $h$ must coincide.
@Willie Wong: I am almost completely sure (eg. Rudin) that the property you describe with respect to real analytic functions, actually applies ONLY to holomorphic functions? Which are generally more 'rigid' than real analytic functions, i.e. they adhere to more rules.