Answer: No this is not true.
At least if I am not mistaken, then $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$ is a counter example for $m=5$. In this case we have: $$\begin{align} U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) \\ U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right) \\ U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1 \end{array}\right) \\ U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2 \end{array}\right) \\ U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr} 5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\ 5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4 \end{array}\right) \end{align}$$
Each $U(k)$ has only $14$ entries $\geq k$.