Let $L^+$ be a set of all real valued functions defined on a real line which are Lebesgue integrable on each $[0,c]$, where $c>0$,
and are zero for $x<0$.
Let for $a>0$, $f_a(t)=(t-a)^{-3/4}$ for $t>a$ and $0$ for $t \leq a$. Then, for $a,b>0$, convolution
$(f_a*f_b)(x):=\int_0^x f_a(x-y)f_b(y)dy=\beta (x-a-b)^{-\frac{1}{2}}$ for $x> a+b$ and $0$ for $x \leq a+b$.
, where $\beta=B(\frac{1}{4}, \frac{1}{4})$.
It shows that convolution of two functions from $L^+$ need not be continuous. Is it possible, maybe by condensation
of singularities and above example to show existence of two functions from $L^+$ which convolution is discontinuous everywhere
on $[0, \infty)$?
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Can convolution on $R_+$ be discontinuous everywhere on $R_+$ ? |
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Let $L^+$ be a set of all real valued functions defined on a real line which are Lebesgue integrable on each $[0,c]$, where $c>0$,
and are zero for $x<0$.
Let for $a>0$, $f_a(t)=(t-a)^{-3/4}$ for $t>a$ and $0$ for $t \leq a$. Then, for $a,b>0$, convolution
$(f_a*f_b)(x):=\int_0^x f_a(x-y)f_b(y)dy=f_{a+b}(x)$f_a(x-y)f_b(y)dy=\beta (x-a-b)^{-\frac{1}{2}}$ for $x> a+b$ and $0$ for $x \leq a+b$.
It shows that convolution of two functions from $L^+$ need not be continuous. Is it possible, maybe by condensation
of singularities and above example to show existence of two functions from $L^+$ which convolution is discontinuous everywhere |
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