Exponential type function

Question

Let $f$ is a function from $\mathbb{R}$ to $\mathbb{C}$ which is normalized function of bounded variation.Let $F(z)$ is another entire function of exponential type $C$.Then show that convolution $f*F(z)$ has exponential type atmost $C$.

Is the following function bounded ??

$\int_{-\infty}^{\infty} |f(z)| \exp(-(C+z)|z|)dz$ It can be used to show above.

Thanks in advance

Answer 1

The integral you have defined need not be finite: if $f(x)=\frac 1 x $ for $x<-1$ and $-1$ for $x \geq -1$ then $f$ is a normalized function of bounded variation but the integral over $(-\infty, -1)$ is $\infty$

Answer 2

To add a little emphasis to the other answer: You cannot show that convolution has exponential type, because the convolution need not even exist. For example take $f=1$, $F=1$.

Maybe you left out a hypothesis?