Length of substring if we just consider a subdivision in $\log n$ substrings

Question

Let $u$ be a string of length $n$ and consider a subdivision in $\log n$ substrings $u = u_1 u_2 \cdots u_{\log n}$. Is it true that there exists a constant $C$ such that for each $1 \le i \le \log n$ the string $$ u_1 u_2 \cdots u_{i-1} u_{i+1} \cdots u_{\log n} $$ has length smaller then $\frac{\log n - 1}{\log n}\cdot n + C$, i.e. $$ \left| u_1 u_2 \cdots u_{i-1} u_{i+1} \cdots u_{\log n} \right| \le \frac{\log n - 1}{\log n}\cdot n + O(1)? $$

Answer 1

Note it is impossible to have all your substrings have length greater than $n / \log n$ So let's make a very mild assumption, let's assume that for increasing $n$ we have our strings $u_i$ such that the length of $u_1$ is $o(n / \log n)$. This means $\lim_{n \to \infty} |u_1| / (n / \log n) = 0$. Note your condition is equivalent to the length, when we remove $u_1$, being $\leq n - n/\log n + C$. And the length when you remove $u_1$ is $n - |u_1|$. Subtracting $n$ gives $|u_1| \geq n / \log n - C$, as $n$ increases. But this contradicts $|u_1| = o(n / \log n)$.