# approximating a discrete function with a continuous one

Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuously differentiable function that reaches a global maximum at $x^*\in(0,1)$. Now, consider its 'discrete' counterpart. That is, consider the collection $\{(x_1,f(x_1)),(x_2,f(x_2)),\ldots,(x_n,f(x_n)\}$ where $x_1<x_2<\cdots <x_n$, and $h=x_{n+1}-x_{n}$ 'small'.

Under what conditions on $h$ (or something else) can I claim that the maximum found using the continuous function $f$ approximates reasonably well the value $f(\hat{x})$ satisfying $f(\hat{x})\geq f(x_i)$ for all $x_i\in \{x_1,x_2,\ldots,x_n\}$?

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I don't understand what $\hat{x}$ is. If I assume that it is it one of the $x_i$, then roughly the error is related to the bound on the derivative of $f$. – user58200 Jan 14 '13 at 4:16
@user58200: Yup. $\hat{x}$ is supposed to be one of the $x_i$'s in the grid. Would you mind elaborating on the argument about the error being related to the bound of the derivative of $f$? Thanks! – Cristian Jan 14 '13 at 4:17
@Cristian: blah essentially answers this. $| \sup{f} - f(\hat{x}) |$ is bounded by $h \sup | f' |$. – user58200 Jan 14 '13 at 4:36

This depends on "how continuous" $f$ is. It's possible to construct continuously differentiable functions that have arbitrarily narrow spikes (e.g. Gaussian as $\sigma \rightarrow 0$). For these functions, sampling at $h$ sized intervals can be arbitrarily wrong.
Using the $\epsilon$-$\delta$ definition of continuity, for any error tolerance $\epsilon > 0$ there exists $\delta > 0$ such that when $|x - \hat x| < \delta$, we have $|f(x) - f(\hat x)| < \epsilon$. In other words, you need to know about $f$ to determine a bound for $h$.
Using differentiability and the mean value theorem, we can deduce that $|f(x) - f(\hat x)| \le |x - \hat x| \sup_{x \in (0, 1)} |f'(x)|$.
I see, but then if the distance $x_{n+1}-x_{n}$ is small (though not infinitesimal), finding the value $x^*_i$ such that $f(x^*_i)>f(x_i)$ for all $x_i$ in the grid can be done using the continuous approximation, correct? – Cristian Jan 14 '13 at 4:28
Right, as long as $f$ is sufficiently well-behaved. If $f$ looks like a smooth interpolation of the discrete $f$, you should be fine. You might run into an off-by-one error where, for example, $f(0.1)$ is the maximum with $h = 0.1$, but $f(0.151)$ is the continuous maximum. Then you might round $0.151$ to $0.2$ and incorrectly conclude that $f(0.2)$ is the discrete maximum. – blah Jan 14 '13 at 4:44