I have a problem understanding the meaning of the delta-function on the sense of distributions. E.g. I have the following equation:

$$\left(\frac{d}{dt} \theta(t) \right) f(t) = \delta(t) f(t)$$

What does this mean for the function $f(t)$? I cannot really understand what the delta function does here in the sense of a distribution.

  • $\begingroup$ A "distribution" is a functional that assigns a number to every function. The "delta function" assigns the number f(0) to the function f. $\delta(t)f(t)= f(0)$ $\endgroup$ – user247327 Dec 11 '15 at 13:57
  • $\begingroup$ Thanks! Actually I need to show that $\delta(t) f(t) = \delta(t)$. I know, that f(0) = 1, so I think that this is correct then? Or does the $\delta(t)$ disappear then after acting on f(t)? $\endgroup$ – Darius Dec 11 '15 at 14:05

I assume $f$ is $\mathcal{C}^{\infty}(\mathbb{R})$.

$\delta f$ is a nonsense, however $f\delta$ has a meaning. One is only allowed to multiply a distribution by a smooth function ($\mathcal{C}^{\infty}(\mathbb{R})\times\mathcal{D}'(\mathbb{R})$), the product of a smooth function by a distribution is not defined.

Definition. Let $T\in\mathcal{D}'(\mathbb{R})$ and $f\in\mathcal{C}^{\infty}(\mathbb{R})$, one define $fT$ by: $$\forall\varphi\in\mathcal{C}^{\infty}_0(\mathbb{R}),\langle fT,\varphi\rangle=\langle T,f\varphi\rangle.$$

Remark. $f\varphi\in\mathcal{C}^{\infty}_0(\mathbb{R})$, since $\textrm{supp}(f\varphi)\subseteq\textrm{supp}(\varphi)$.

One has the:

Proposition. Let $T\in\mathcal{D}'(\mathbb{R})$ and $f\in\mathcal{C}^{\infty}(\mathbb{R})$, $fT\in\mathcal{D}'(\mathbb{R})$.

Proof. Since $T$ is a distribution, $fT$ is cleary linear map from $\mathcal{C}^{\infty}_0(\mathbb{R})$ to $\mathcal{C}^{\infty}_0(\mathbb{R})$. Let $K$ be a compact of $\mathbb{R}$, since, $T$ is a distribution there exists $C\in\mathbb{R}_+^*$ and $k\in\mathbb{N}$ such that: $$\forall\varphi\in\mathcal{C}_0^{\infty}(\mathbb{R}),\textrm{supp}(\varphi)\subseteq K,|\langle T,\varphi\rangle|\leqslant C\sum_{i=1}^n\sup_{K}\left|\varphi^{(i)}\right| $$ Let $\varphi\in\mathcal{C}^{\infty}_0(\mathbb{R})$, $\textrm{supp}(f\varphi)\subset K$ and one has: $$\begin{align}|\langle fT,\varphi\rangle|&=|\langle T,f\varphi\rangle|\\&\leqslant C\sum_{i=1}^k\sup_{K}\left|(f\varphi)^{(i)}\right|\\&\leqslant C\sum_{i=1}^k\sup_{K}\left|\sum_{j=0}^i{i\choose j}f^{(j)}\varphi^{(i-j)}\right|\\&\leqslant C\sum_{i=1}^k\sum_{j=0}^i{i\choose j}\sup_{K}\left|f^{(i)}\right|\sup_{K}\left|\varphi^{(j-i)}\right|\\&\leqslant\widetilde{C}\sum_{i=1}^k\sup_{K}\left|\varphi^{(i)}\right|\end{align}.$$ Hence, $fT$ is a distribution. $\Box$

Remark. $\widetilde{C}$ can for example be : $$C\max_{i\in\{1,\ldots,k\}}\left(\max_{j\in\{1,\ldots,i\}}{i\choose j}\sup_{K}\left|f^{(i)}\right|\right).$$

In your case, one has: $$\forall\varphi\in\mathcal{C}_0^{\infty}(\mathbb{R}),\langle f\delta,\varphi\rangle=(f\varphi)(0)=f(0)\langle\delta,\varphi\rangle.$$ Therefore, one gets: $$f\delta=f(0)\delta.$$

  • $\begingroup$ This is the best answer I could ever think of. Thank you so much! And you are right, it was of course $f \delta$ and not $\delta f$. $\endgroup$ – Darius Dec 12 '15 at 8:41

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