We consider the function $f$ defined by $$ f(x) = \begin{cases} \dfrac{\sin x}{2} &: x \in ]-\infty,\dfrac{\pi}{2}[\\ 0 &: x \in [\dfrac{\pi}{2},+\infty[ \end{cases} $$ The questions are:

  1. calculate $(T_f)'$,

  2. calculate $(T_f)''$,

  3. deduce the differential equation satisfied by $(T_f)'$ and $(T_f)''$.

My solution is:

$f \in L^1_{loc}(\mathbb{R})$, so we can define the distribution $T_f$ by the relation

$$ \forall \varphi \in \mathcal{D}(\mathbb{R}), \langle T_f,\varphi \rangle = \displaystyle\int_{-\infty}^{+\infty} f(x) \varphi(x) dx= \displaystyle\int_{-\infty}^{\pi/2} \dfrac{\sin x}{2} \varphi (x) dx. $$

  1. Calclulate $(T_f)'$: we have $$ (T_f)'= T_{f'} + (f(\dfrac{\pi}{2}^+) - f(\dfrac{\pi}{2}^-)) \delta_{\pi/2} = \dfrac{1}{2} \cos(x) + \delta_{\pi/2}. $$

My problem is: How I can calculate $(T_f)''$? with an rule for example, and how we deduce the differential equation satisfied by $(T_f)'$ and $(T_f)''$? Please.

  • $\begingroup$ What is $T$???? $\endgroup$
    – zoli
    Mar 3, 2017 at 12:10
  • $\begingroup$ i edit my first post. Thank you for advice. $\endgroup$
    – user415040
    Mar 3, 2017 at 12:32
  • $\begingroup$ There are two problems with your $T'_f$: first, you forgot that $\frac \sin x$ was used only for $x<\pi/2$, second, the coefficient at $\delta_{\pi/2}$ is not $1$. $\endgroup$ Mar 3, 2017 at 13:03
  • $\begingroup$ Sorry but i don't understand. $\endgroup$
    – user415040
    Mar 3, 2017 at 13:06

1 Answer 1


$$ \langle T',\phi\rangle=-\langle T,\phi'\rangle=-\frac12\int_{\pi/2}^\infty(\sin x)\,\phi'(x)\,dx=\frac12\int_{\pi/2}^\infty(\cos x)\,\phi(x)\,dx+\frac12\,\phi(\pi/2). $$ Now for the second derivative: $$\begin{align} \langle T'',\phi\rangle&=-\langle T',\phi'\rangle&\\ &=-\frac12\int_{\pi/2}^\infty(\cos x)\,\phi'(x)\,dx-\frac12\,\phi'(\pi/2)\\ &=-\frac12\int_{\pi/2}^\infty(\sin x)\,\phi(x)\,dx-\frac12\,\phi'(\pi/2). \end{align}$$ Thus $T''=-T+\dfrac12\,\delta_{\pi/2}'$.


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