Exercise 9.28

Away from the origin, $\phi$ is continuous since the definitions agree at the points $(x,0)$, $(x,\sqrt{t})$ and $(x,2\sqrt{t})$. Since $|\phi (x,t)|\le \sqrt{|t|}$, we have for $𝜀>0$, $-𝜀<x<𝜀$, and $-𝜀<t<𝜀$, $|\phi (x,t)|<\sqrt{𝜀}$, so $\phi$ is also continuous at the origin.

For $x\le 0$ we have $(x,t)=0$ for all $t$, and for $x>0$ we have $(x,t)=0$ for $-\frac{1}{2}{x}^2\le t\le \frac{1}{2}{x}^2$, so $D_2(x,0)=0$ for all $x$.

If $0\le t<\frac{1}{4}$, then

and if $-\frac{1}{4}<t\le 0$, then

Hence $f^{\prime }(0)=1$, while ${\int }_{-1}^1{D}_2\phi (x,0)\mathit{dx}=0$.