Sign of the second order derivative in the sense of distributions at a point of discontinuity

Question

Let $f:[a-1,1] \to \mathbb R_{+}$ such that $f(x)=\begin{cases} 1-(a-x)^2 &a-1 \lt x \leq a+1 \\ 0 & a+1 < x \leq 1 \end{cases}$

for some real $a<0$.

If I compute the first order derivative of $f\;$, I obtain $f^{\prime}(x)=\begin{cases} 2(a-x) &a-1 \lt x \leq a+1 \\ 0 & a+1 < x \leq 1 \end{cases} \quad$ which has a discontinuity of first kind at $x=a+1$.

I am interested in estimating the sign of the second order derivative at $x=a+1$. To this end, I tried to compute this derivative in the sense of distributions. If $\phi \in \mathcal C^1_c([a-1,1])$ then

$\begin{align} \frac{d}{dx} f^{\prime}(\phi)=-\int_{a-1}^{a+1} 2(a-x)\phi^{\prime}(x)\;dx&=2\phi(a+1)+2\int_{a-1}^{a+1} \phi(x)\;dx\\&=2\delta_{a+1}(\phi)+2\int_{a-1}^{a+1} \phi(x)\;dx \end{align}$

At this point, I got stuck! Are my computations correct? If yes, then what am I missing? I don't see which will the sign of the second derivative of $f$ be at $x=a+1$... Any help is much appreciated.

Thanks in advance!

Answer 1

To find the distributional equivalent of $f''$, one calculates, say $\varphi\,' \in \mathcal{C^\infty_c([a-1,1])}$, and calculates the Schwartz bracket quantity, $$ \begin{align} \langle f'',\,\varphi'\rangle &= - \langle f',\,\varphi''\rangle \\ &= -\int\limits_{a-1}^1 f'(x)\,\varphi''(x)\;{\rm d} x \\ &= -2\int\limits_{a-1}^{a+1} (a-x)\,\varphi''(x)\;{\rm d} x -\int\limits_{a+1}^1 0\cdot\,\varphi''(x)\;{\rm d} x \\ &= -2\int\limits_{a-1}^{a+1} (a-x)\,\varphi''(x)\;{\rm d} x \\ &= -2a\int\limits_{a-1}^{a+1}\varphi''(x)\;{\rm d} x + 2\int\limits_{a-1}^{a+1} x\,\varphi''(x)\;{\rm d} x \\ &= 2\varphi'(a+1) - 2\varphi(a+1) \\ &= 2\langle(\delta_{a+1},\,\varphi'\rangle + 2\langle(h_{a+1},\,\varphi'\rangle \\ &= \langle f'',\,\varphi'\rangle. \end{align} $$ The above is true for all $\varphi'$ in the test function space, and so therefore one must conclude from the last two lines that, $$ f'' \doteq 2\delta_{a+1} + 2h_{a+1}, $$ where the symbol $\doteq$ means "is equal in the sense of distributions", and $h(x)$ is the Heaviside function which is $1$ for $x > 0$ and $0$ for $x<0$. $\qquad\blacksquare$

Answer 2

You have $f'((a+1)^-) = -2$ and $f'((a+1)^+) = 0$ so $f'$ increases at $x=a+1$. Therefore the sign of $f''$ at $x=a+1$ is positive.