Since the Fourier Transform of the product of two functions is the same as the convolution of their Fourier Transforms, and the Fourier Transform is an isometry on $L^2$, all we need find is an $L^2$ function that when squared is no longer an $L^2$ function.
Transforms such as Fourier transform or Laplace transform, takes a product of two functions to the convolution of the integral transforms, and vice versa. This is called the Convolution Theorem, and is available with proof at wikipedia.
Lowercase t-like symbol is a greek letter "tau". Here it represents an integration (dummy) variable, which "runs" from lower integration limit, "0", to upper integration limit, "t". So, the convolution is a function, which value for any value of argument (independent variable) "t" is expressed as an integral over dummy variable "tau".
But we can still find valid Laplace transforms of f (t) = t and g (t) = (t^2). If we multiply their Laplace transforms, and then inverse Laplace transform the result, shouldn't the result be a convolution of f and g?
1 The purpose of this answer is to show how a direct application of convolution may lead to the desired result. I take the following results from Cohn, Measure Theory. Definition of convolution Let $\nu_1$ and $\nu_2$ be finite measures on $ (\mathbb {R}^d,\mathscr {B} (\mathbb {R}^d))$, then their convolution $\nu_1\ast\nu_2$ is defined by:
Convolution with sinc pulses What we want to do to reconstruct the signal is a convolution between the samples and scaled and shifted versions of sinc. This technique is known as Whittaker–Shannon interpolation: " This is equivalent to filtering the impulse train with an ideal (brick-wall) low-pass filter with gain of 1 (or 0 dB) in the passband.
