Number of roots of analytic function restricted to $(0,1)$

The motivation for this question is the same as in my previous question in MO: http://mathoverflow.net/questions/115179/real-root-1-of-the-hasse-weil-l-function-of-c-over

Let us consider an analytic function $f$ defined in the whole complex plane which has infinitely many zeros. Let us restrict the function to the interval $(0,1)$ as follow: $g(t)=f(1-2t)$. I look for the number of roots of $g$ in $(0,1)$.

My question is: What I can say for the case of $g$ defined by using $f$ in $(0,1)$.

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What do you mean by "for analytic function, the number is infinite"? There are analytic functions, such as the exponential function, that have no zeros at all. – joriki Dec 15 '12 at 14:04
Only for the function f described here. – ZE1 Dec 15 '12 at 14:07
But you haven't described one? Your edit has very slightly improved things, because it wasn't clear before whether you meant $f$ or $g$, but you still haven't said anything about $f$ that would suggest that it has infinitely many zeros. – joriki Dec 15 '12 at 14:08
The link to the MO question, though certainly helpful, hasn't shed any light on my questions above. – joriki Dec 15 '12 at 14:15
i added this as an edit but it was reverted... comment then: this is a duplicate of mathoverflow.net/questions/116447/… as is easy to find if you click through to the second page of questions on the overflow account for the question, that was linked. – Peter Sheldrick Dec 15 '12 at 15:46

The number of zeros of $g$ in $(0,1)$ is equal to the number of zeros of $f$ in $(-1,1)$. If you don't know where the infinitely many zeros of $f$ are, you don't know anything about the zeros of $g$.