CF 103107L - Labi-Ribi
Fix an integer $t ge 1$. Let $N ge 0$ be given. Define $kappat N$ in the discrete sense (as in earlier parts of Section 7.2.1.3) as the unique integer $m ge t-1$ such that $$binom{m}{t} le N < binom{m+1}{t},$$ and set $$kappat N = binom{m}{t-1}.
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Solution
Setup
Fix an integer $t \ge 1$. Let $N \ge 0$ be given. Define $\kappa_t N$ in the discrete sense (as in earlier parts of Section 7.2.1.3) as the unique integer $m \ge t-1$ such that
$$\binom{m}{t} \le N < \binom{m+1}{t},$$
and set
$$\kappa_t N = \binom{m}{t-1}.$$
Now define the continuous extension as follows. For $x \ge t-1$, the function $x \mapsto \binom{x}{t}$ is strictly increasing, hence invertible onto $[0,\infty)$. For each $N \ge 0$, let $x \ge t-1$ satisfy
$$N = \binom{x}{t},$$
and define
$$\widetilde{\kappa}_t N = \binom{x}{t-1}.$$
The goal is to prove
$$\kappa_t N \le \widetilde{\kappa}_t N$$
for all integers $t \ge 1$ and $N \ge 0$.
Solution
Let $N \ge 0$ and choose $x \ge t-1$ such that $N = \binom{x}{t}$. Let $m$ be the integer determined by
$$\binom{m}{t} \le \binom{x}{t} < \binom{m+1}{t}.$$
Since $x \mapsto \binom{x}{t}$ is strictly increasing on $[t-1,\infty)$, the inequalities imply $m \le x < m+1$.
The function $x \mapsto \binom{x}{t-1}$ is also strictly increasing on $[t-2,\infty)$. For real $x \ge m \ge t-1$, the monotonicity yields
$$\binom{m}{t-1} \le \binom{x}{t-1}.$$
By the definition of $\kappa_t N$ in the discrete sense,
$$\kappa_t N = \binom{m}{t-1}.$$
By the definition of the continuous extension,
$$\widetilde{\kappa}_t N = \binom{x}{t-1}.$$
Substitution into the inequality gives
$$\kappa_t N \le \widetilde{\kappa}_t N.$$
This completes the proof. ∎
Verification
For real $x \ge t-1$, the expression
$$\binom{x}{t} = \frac{x(x-1)\cdots(x-t+1)}{t!}$$
is a product of $t$ linear factors, each nondecreasing in $x$ on $[t-1,\infty)$, hence the product is strictly increasing. The same argument applies to $\binom{x}{t-1}$ on $[t-2,\infty)$.
If $m$ is defined by $\binom{m}{t} \le \binom{x}{t} < \binom{m+1}{t}$, monotonicity forces $m \le x < m+1$, since otherwise the strict increase would contradict the ordering.
The inequality $\binom{m}{t-1} \le \binom{x}{t-1}$ follows directly from monotonicity with $m \le x$.
All substitutions are consistent with the definitions of $\kappa_t N$ and $\widetilde{\kappa}_t N$.
Notes
The argument relies only on monotonicity of generalized binomial coefficients on $[t-1,\infty)$ and does not require convexity or differentiability. Equality holds when $x$ is an integer because then $N = \binom{x}{t}$ forces $m = x$, so both definitions give $\binom{x}{t-1}$.