Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$.

Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $a(1)=1$ and $a(x)=a(-x)\forall x$.

Suppose that we have a convex function $f:[-2,2]\to\mathbb{R}$ such that $f''\leq0$, $(af)''\leq0$. We can also suppose that $f'\leq0$ in $[0,2]$, if not change $f(x)$ by $f(-x)$. So $f(0)\geq f(1)$. Now apply Jensen's inequality to $af$ to obtain $f(1)=(af)(1)\geq\frac{1}{2}((af)(0)+(af)(2))\geq\frac{1}{2}(af)(0)=\frac{3}{2}f(0)\geq\frac{3}{2}f(1)>f(1)$, a contradiction.

Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$.

Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $a(1)=1$ and $a(x)=a(-x)\forall x$.

Suppose that we have a convex function $f:[-2,2]\to\mathbb{R}$ such that $f''\leq0$, $(af)''\leq0$. We can assume $f'(0)\leq0$ (if not change $f(x)$ by $f(-x)$). So $f$ is decreasing in $[0,2]$ and we have $f(0)\geq f(1)$. Now apply Jensen's inequality to $af$ to obtain a contradiction:

$$f(1)=(af)(1)\geq\frac{1}{2}((af)(0)+(af)(2))\geq\frac{1}{2}(af)(0)=\frac{3}{2}f(0)\geq\frac{3}{2}f(1)>f(1)$$