[$]\phi\big(\frac{x}{\sqrt t}) = \frac1{\sqrt{2\pi}}\exp\big(-\frac{x^2}{2t}\big)[$] with measure [$]dx[$]. Do keep in mind, as I have said in my last post, you can replace [$]\sqrt t[$] with [$]t^k[$] for a corresponding positive [$]k[$] to achieve any desired convergence speed.[$]\phi\big(\frac{x}{\sqrt t}) = exp(-x^2/4t)[$], Yes?Late to the party.

I like Alan's heat equation idea immensely. In fact the Fourier solution and the image solution are just the two (dual) sides of the Poisson summation formula converging faster at [$]t\searrow 0[$] and [$]t\rightarrow\infty[$] respectively. One possible drawback of the [$]\frac1t[$] solution is the integration of the Green's function has to be enumerated either as a sum of error functions or numerically. Both the Chebyshev polynomial series method of outrun and Fourier series method of ppauper are good, but, just as outrun has pointed out, they converge slowly. The reason is, just as outrun suspected, due to the discontinuity in the derivative of the original function. The convergence speed is [$]O(\frac1n)[$] or even lower, where [$]n[$] is the index of the series. Also the derivative close to the vertex is not uniformly convergent but osciallates around the discontinuity. This is called the Gibbs phenomenon. All of these approximations are actually analytical functions (as in their Taylor power series are convergent), not only infinitely differentiable.

Here are 2 more approximations satisfying the requirement.

1. A silly one that is not analytic (Taylor power series is undefined at two points) but infinitely differentiable that does satisfy the requirement. For the sake of convenience I linear transform the Cuchulainn's hat function into the absolute function [$]|x|[$]. My approximation is simply [$]|x|+e^{\frac{b}{(x+a)(x-a)}}\Theta(a^2-x^2)[$] where [$]a>0[$] and [$]b>0[$] are parameters to be adjusted according to your requirement of the fidelity of the approximation, and [$]\Theta[$] is the Heaviside function. The trajectory of my approximating function simply rounds off the sharp corner near it and assumes the shape of original function away from the corner with the whole curve being infinitely differentiable.

2. Any smoothing summation kernel, convolution or not, would work. The smoothness (differentiability or analycity) of the resulting function is determined by that of the kernel. The convergence can be controlled by a parameter. It is then the Dirac delta functional. A simple example of this is simply the convolution operator [$]\Phi[t,f]:=\frac1{\sqrt t}\phi\big(\frac{\cdot}{\sqrt t}\big)*f(\cdot)[$], where [$]\phi[$] is, for example, the standard Gaussian function. The convolution is analytic since [$]\phi[$] is and the convolution uniformly converges with high speed to [$]f[$] as [$]t\rightarrow0[$]. This example is just the heat equation solution with the initial condition extended by the zero function to infinity, dispensing with the finite boundary condition and thus the infinite source images for the Green's function, or reflection and shifting of [$]f[$].

So, the motivation is to solve heat equation in an infinite rod and the solution is a convolution integral between this kernel and [$]f[$]? I just need to get to the stage to roll out a solution which is what Alan sketched? I need to check those steps,

Your solution is a kind of stroke of genius.

Yes to all your questions, except (just to state the obvious to be safe) to emphasize that we extend the wedge shape of [$]f[$] to infinity for the convolution with [$]\phi[$]. Basically, it is the shape (after linear transformation) of [$]|x|[$]. Please do check Alan's expression. It has the right look but I have not checked the exact computation.

Do you like my solution #1? I thought it was even better. What may be your reservation?

Oh, you flatter me. Thank you.