Let $\eta : \mathbb{R} \to \mathbb{R}_0^+$ be a function of class $C_0^2$ such that $\eta(t) \geq 1$ for $|t| \leq 1/2$. Here $\mathbb{R}_0^+$ denotes the set of all positive real numbers including zero.
I am wondering what is the smallest possible value that the product $\|\eta\|_1 \|\eta''\|_1$ can be? Here $\|f\|_1 := \int_{\mathbb{R}}|f(x)|dx$. Can we find such an $\eta$ so that the product is minimized?
Let \begin{align*} M_{+} & =\max\{|\eta^{\prime}(x)|:\,x\geq0\}=|\eta^{\prime}(x_{+})|,\\ M_{-} & =\max\{|\eta^{\prime}(x)|:\,x\leq0\}=|\eta^{\prime}(x_{-})|. \end{align*} Since $\eta$ has compact support, there exist $x_{1}<x_{-}$ and $x_{2}>x_{+}$ such that $\eta^{\prime}(x_{1})=\eta^{\prime}(x_{2})=0$. By the fundamental theorem of calculus, \begin{align*} M_{+} & =|\eta^{\prime}(x_{+})|=|\eta^{\prime}(x_{+})-\eta^{\prime}% (x_{2})|\leq\int_{x_{+}}^{x_{2}}|\eta^{\prime\prime}(t)|\,dt,\\ M_{-} & =|\eta^{\prime}(x_{-})|=|\eta^{\prime}(x_{-})-\eta^{\prime}% (x_{1})|\leq\int_{x_{1}}^{x_{-}}|\eta^{\prime\prime}(t)|\,dt, \end{align*} which shows that $$ M_{+}+M_{-}\leq\Vert\eta^{\prime\prime}\Vert_{L^{1}}. $$ Since $\eta(0)\geq1$ and $\eta$ has compact support, there exists a first time $x_{3}>0$ such that $\eta(x_{3})=\frac{1}{2}$ and so by the fundamental theorem of calculus, $$ \frac{1}{2}\leq\eta(0)-\eta(x_{3})\leq\int_{0}^{x_{3}}|\eta^{\prime }(t)|\,dt\leq M_{+}(x_{3}-0), $$ which implies that $x_{3}\geq\frac{1}{2M_{+}}$. But then $$ \int_{0}^{x_{3}}\eta(t)\,dt\geq\frac{1}{2}x_{3}\geq\frac{1}{4M_{+}}. $$ Similarly on the negative side$$ \int_{x_{4}}^{0}\eta(t)\,dt\geq\frac{1}{4M_{-}}, $$ where $x_{4}<0$ is the last negative time such that $\eta$ is $\frac{1}{2}$. This shows that$$ \frac{1}{4M_{+}}+\frac{1}{4M_{-}}_{-}\leq\Vert\eta\Vert_{L^{1}}. $$ Hence,$$ \Vert\eta\Vert_{L^{1}}\Vert\eta^{\prime\prime}\Vert_{L^{1}}\geq(M_{+}% +M_{-})\left( \frac{1}{4M_{+}}+\frac{1}{4M_{-}}\right) \geq\frac{1}{2}. $$ These computations are not sharp, but they show that the infimum is positive.