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 $f \in BV(\mathbb R)$?
A general principle is that the regularity (degree of differentiability) of a convolution $f*g$ will be the sum of the regularities of $f,g$ separately, but no better (e.g., the convolution of a (compactly supported) $C^2$ function and a $C^3$ function will be $C^5$ but no better). $L^1 \cap L^\infty$ has 0 degrees of regularity and BV has one, so this suggests that the first question has a negative answer and (as a first approximation) the second question could have a positive answer.
To formalise this heuristic one can work with modulated bump functions $f(x) = e^{iNx} \psi(x)$, $K(x) = N^{-1} e^{iNx} \psi(x)$ for some large frequency parameter $N \geq 1$ and some test function $\psi$ (one could also take a gaussian if one prefers). Note that $f$ is bounded in $L^1 \cap L^\infty$ and $K$ is bounded in $BV$, but the convolution $$ f*K(x) = N^{-1} e^{iNx} (\psi * \psi)(x)$$ has a second derivative that grows like $N$ in $L^1$ norm. (The negative powers of $N$ here correspond to the degrees of differentiability in the previous heuristic.) By taking suitable convergent linear combinations of this example one can find $f,K$ that are bounded in $L^1 \cap L^\infty$, $K$ respectively but for which $(f*K)''$ fails to be in $L^1$. (One may possibly also be able to reach this conclusion via some bilinear version of the closed graph theorem, though I do not know offhand of a suitable such variant of that theorem.)
The modulated bump functions no longer provide a counterexample when $f$ is required to lie in $BV$, as now an additional factor of $N^{-1}$ must be placed in front of $f$ and the convolution $f*K$ now acquires a factor of $N^{-2}$ instead of $N^{-1}$. And indeed the claim is close to being true: if $f,K$ are in $BV$ then $f',K'$ are finite measures and $(f*K)'' = f'*K'$ will also be a finite measure, which is not that much weaker than being in $L^1$. However that's the best one can get, as Christian's example already shows (take e.g., $f = 1_{[0,1]}$ and $K = 1_{[0,\infty]}$, then $(f*K)'' = \delta_0 - \delta_1$. However, if one upgades one of $f,K$ from $BV$ to $W^{1,1}$ then $(f*K)''$ will similarly be upgraded to $L^1$.
Incidentally it may be helpful to view all the standard function spaces on a two-dimensional diagram of regularity and integrability (sometimes known as a de Vore diagram), see e.g., this blog post of mine. To incorporate integrability into the above heuristic one can replace the bump functions $\psi(x)$ by rescaled bump functions $\psi(x/\varepsilon)$ for various $\varepsilon>0$ (usually restricting to the regime $\varepsilon \geq 1/N$ to respect the uncertainty principle). Roughly speaking, one then arrives at the conclusion that the regularity and integrability of a convolution is (as a first approximation) basically what one would expect by adding up the regularities of the factors, and then applying Young's inequality, and (optionally) Sobolev embedding to trade some of the regularity for integrability.
