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Hiro
Hiro
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Is $(f \ast K)'' \in L^1(\mathbb R)$ wherefor $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$?

Is it possible to deduce that $$(f \ast K)'' \in L^1(\mathbb R)$$ if $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$? What I can prove is that $(f \ast K)' \in L^1 \cap L^\infty$. Is the bound on the second derivative also true?

If the above does not suffice, is the result true with the additional assumption $f \in BV(\mathbb R)$?

Is $(f \ast K)'' \in L^1(\mathbb R)$ where $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$

Is it possible to deduce that $$(f \ast K)'' \in L^1(\mathbb R)$$ if $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$? What I can prove is that $(f \ast K)' \in L^1 \cap L^\infty$. Is the bound on the second derivative also true?

Is $(f \ast K)'' \in L^1(\mathbb R)$ for $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$?

Is it possible to deduce that $$(f \ast K)'' \in L^1(\mathbb R)$$ if $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$? What I can prove is that $(f \ast K)' \in L^1 \cap L^\infty$. Is the bound on the second derivative also true?

If the above does not suffice, is the result true with the additional assumption $f \in BV(\mathbb R)$?

Source Link
Hiro
Hiro
  • 131
  • 6

Is $(f \ast K)'' \in L^1(\mathbb R)$ where $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$

Is it possible to deduce that $$(f \ast K)'' \in L^1(\mathbb R)$$ if $f \in L^1 \cap L^\infty(\mathbb R)$ and $K \in BV(\mathbb R)$? What I can prove is that $(f \ast K)' \in L^1 \cap L^\infty$. Is the bound on the second derivative also true?