= Solution
The constant is sharp because continuous functions bounded by one can approximate $\operatorname{sgn}K$ arbitrarily closely outside an arbitrarily small interval around zero. Choose $f'''$ to be such an approximation and integrate three times to obtain $f\in C^3[-1,1]$. Then
$$
\frac{|L(f)|}{\|f'''\|_\infty}
=\frac{\left|\int K(t)f'''(t)\,dt\right|}
{\|f'''\|_\infty}
\longrightarrow\int_{-1}^1|K(t)|\,dt=\frac13.
$$
\b[No smaller constant can therefore satisfy the inequality for every $f$.]
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