Approximation using a Fourier transform with low pass filter
Depends on what you mean by 'best'.  In terms of least squares, then yes, just knocking off the terms above 1kHz will give the closest approximation.  I guess one way to prove it would be:
$$
     x(t) = \int_{-\infty}^{\infty} X(f)\exp\left( j2\pi ft \right) dt\\
\hat x(t) = \int_{-1k}^{1k} X(f)        \exp\left( j2\pi ft \right) dt
$$
leaving a residual error of
$$
e(t) = \int_{|f| &gt; 1k} X(f)\exp\left( j2\pi ft \right) dt .
$$

Compare this to a 'fiddled' version, using $\tilde X(f)$ for the coefficients:
$$
\hat x(t) = \int_{-1k}^{1k} \tilde X(f) \exp\left( j2\pi ft \right) dt,
$$
which has a residual error of
$$
e(t) = \int_{|f| \le 1k} \left(X(f) - \tilde X(f)\right) \exp\left( j2\pi ft \right) dt + 
\int_{|f| &gt; 1k} X(f)\exp\left( j2\pi ft \right) dt ,
$$
which is greater than the error in the first instance unless $\tilde X(f) = X(f) \; \forall f$.

Something like that ought to do it, anyway.

EDIT: It is possible to prove that a projection onto any orthogonal set of functions (doesn't need to be the Fourier set) gives the least squares fit:

$$
\int f_k(t) f_m(t) dt = \cases{
  0, &amp; $k \ne m$ \\
  1/\lambda_k, &amp; $k = m$}
$$
for any orthogonal basis set $\{f_k\}$ (I'm not bothering with weighting functions), where
$$
\lambda_k = \frac{1}{\int f^2(t) dt}
$$
The projection of $g(t)$ onto the set is given by
$$
c_k = \lambda_k \int g(t)f_k(t) dt.
$$
The least squares fit of a set of orthogonal basis function is the minimisation of
$$
\int \left[g(t) - \sum_k d_k f_k(t)\right]^2 dt.
$$
Minimise by setting the derivative (wrt $d_m$) to zero:
$$
2\int\left(g(t) - \sum_kd_kf_k(t)\right)\left(-f_m(t)\right) dt = 0 \\
-\int f_m(t)g(t)dt + \int f_m(t)\sum_kd_kf_k(t)dt = 0 \\
-\int f_m(t)g(t)dt + \sum_kd_k \int f_m(t)f_k(t)dt = 0 \\
-\int f_m(t)g(t)dt + \sum_kd_k \frac{1}{\lambda_m} = 0
$$
which rearranges to give the same value as for $c_k$, thus the least squares fit is exactly the straightforward projection onto the orthogonal set.