Given an even function $f(x)$, how to obtain the most close to it continuous odd function $g(x)$?

By most close I mean that $\int_0^\infty |f(x)-g(x)| dx$ be the minimum possible and the difference $|f(x)-g(x)|$ be monotonously decreasing for x>0.

To avoid trivial solutions suggested by James Cranch, let postulate for simplicity that the both functions are discrete-analytic, that is equal to their Newton expansions.

$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$

$$g(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k g\left (0\right)$$

Or in the weaker case, bi-directional Newton expansion

$$f(x) = \lim_{n\to\infty}\sum_{k=-n}^n \binom{x}k \Delta^k f\left (n\right)$$

$$g(x) = \lim_{n\to\infty} \sum_{k=-n}^n \binom{x}k \Delta^k g\left (n\right)$$

Am I right that hyperbolic sine and cosine are the most close even and odd functions to each other?

I would prefer a plain and simple expression for this transformation operator.

Note that since we can expand the definition of discrete-analytic functions to those having poles in integer points, using this formula

$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k \lim_{t->k}\left( f(t)\prod_{j=0}^\infty(t-x_j)\right)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k\prod_{j=0}^\infty(k-x_j)}{(x-k) (k+n)! (n-k)!}}$$

where $x_j$ are the poles, we can similarly talk about such most close even and odd functions even if they have poles:

The motivation for this question is as follows: it seems that this relation connects elementary functions to zeta function and differentiated gamma functions. Particularly, the most close odd function to the even elementary function $f(x)=\csc^2 x - \frac1{x^2}$ is not elementary.

On the graphic above, the blue function is elementary while similarly looking red one is not!

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