We have a functional over $\mathbb {R} \to \mathbb {R}$ functions $\{f\}$, that could be written as

$F (f) := \int_{-\infty}^\infty \mathbb{d}t K (t) f (t)$,

where $K(t)$ is a distribution that could be a function or a generalized function (such as Dirac delta).

If we wish for the functional to extract the functions's value at $t_0$, we simply take $K(t) = \delta (t - t_0)$.

However, what if we concentrate solely on a subsection of the integration space, i.e. lets define the functional as

$F_+ (f) := \int_{0}^\infty \mathbb{d}t K (t) f (t)$, ($f$ is still well defined on whole $\mathbb{R}$),

we now wish to find again $K (t)$ such that $F_+ (f) = f (t_0)$, for $t_0 > 0$ this is again trivially the Dirac delta, but what about $t_0 < 0$? Is there any general distribution $K (t)$ such that the distribution would give $f (t_0), t_0 < 0$, for an arbitrary function?

If not, is there $K (t)$ with such properties of a specific class of functions?

I.e. if we have a class of functions

$g_{a, b} (t) := \frac {a t} {(a b)^2 + (t - b)^2}$, $\mathbb {R} \to <g_{min}, g_{max}>$, $a, b > 0$,

characterised by 2 positive real parameters, does there exist $K (t)$, possibly dependent on $a$, neccesarily indepentent of $b$, such that

$F_+ (g_{a, b}) := \int_{0}^\infty \mathbb{d}t K_a (t) g_{a, b} (t) = g_{a, b} (t_0), t_0 < 0$?

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    $\begingroup$ There definitely won't be a general formula, since a general function's value at $t=-1$ (say) is completely independent of its values for $t\ge0$. $\endgroup$ Mar 18 at 20:18

1 Answer 1


I'll cheat a little bit : if you allow $K(t)$ to be an operator, as $K(t) = \delta(t-\tau)\,e^{(t_0-\tau)\partial_t}$ for instance, with $\tau > 0$ and $t_0 < 0$, then you can construct some relations as follows : $$ F[f] = \int_0^\infty K(t)f(t) \,\mathrm{d}t = \int_0^\infty \delta(t-\tau)\,e^{(t_0-\tau)\partial_t}f(t) \,\mathrm{d}t = \int_0^\infty \delta(t-\tau)f(t+t_0-\tau) \,\mathrm{d}t = f(t_0) $$


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