Question

Define $g\left(x\right)=\underset{-3}{\overset{3}{\int }}f(x-y)f\left(y\right)dy$, for all real $x$, where
$f\left(t\right)=\{\begin{array}{cc}1& 0\le t\le 1\\ 0& \text{elsewhere}.\end{array}$
Then

• A. $g\left(x\right)$ is not continuous everywhere
• B. $g\left(x\right)$ is continuous everywhere but differentiable nowhere
• C. $g\left(x\right)$ is continuous everywhere and differentiable everywhere except at $x=0,1$
• D. $g\left(x\right)$ is continuous everywhere and differentiable everywhere except at $x=0,1,2$

Answer 1

The correct option is D $g\left(x\right)$ is continuous everywhere and differentiable everywhere except at $x=0,1,2$

$g\left(x\right)=\underset{-3}{\overset{3}{\int }}f(x-y)f\left(y\right)dy\Rightarrow g\left(x\right)=\underset{0}{\overset{1}{\int }}f(x-y)dy$
Assuming $x-y=t\Rightarrow dy=-dt$
$\text{So,}g\left(x\right)=\underset{x-1}{\overset{x}{\int }}f\left(t\right)dt$

$\text{Case I :}x<0$
$f\left(t\right)=0\Rightarrow g\left(x\right)=0$

$\text{Case II :}0\le x<1$
$g\left(x\right)=\underset{x-1}{\overset{0}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow g\left(x\right)=\underset{0}{\overset{x}{\int }}1dt=x$

$\text{Case III :}1\le x\le 2$
$g\left(x\right)=\underset{x-1}{\overset{1}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow g\left(x\right)=\underset{x-1}{\overset{1}{\int }}1dt=2-x$

$\text{Case IV :}x>2$
$f\left(t\right)=0\Rightarrow g\left(x\right)=0$

$\therefore g\left(x\right)=\{\begin{array}{cc}0& x<0\\ x& 0\le x<1\\ 2-x& 1\le x\le 2\\ 0& x>2\end{array}$

Clearly, $g\left(x\right)$ is continuous $\forall x\in R$ and $g\left(x\right)\text{is differentiable}\forall x\in R-\{0,1,2\}.$