Can a function "grow too fast" to be real analytic?

Question

Does there exist a continuous function $\: f : \mathbf{R} \to \mathbf{R} \:$ such that for
all real analytic functions $\: g : \mathbf{R} \to \mathbf{R} \:$, for all real numbers $x$,
there exists a real number $y$ such that $\: x < y \:$ and $\: g(y) < f(y) \:$?

Answer 1

No. Only if you require $g$ or its coefficients to be computable. Suppose there is such an $f$, then we could just pick the points $(n,(1+\sup\{ f(z))|n-1<z<n+1\}))$, for $n=1,2,3\ldots$ and interpolate.

Answer 2

Just take $f(x) = \tan(x)$ (defining $f(x) = 0$, say, when $x$ is an integer multiple of $\pi/2$. But this has nothing to do with "growing too fast".