53. Real numbers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=0$, and $2 a_{k} \leqslant a_{k-1}+$ $a_{k+1}, k=2,3, \cdots, n-1$. Find the smallest $\lambda(n)$, such that for all $k \in\{1,2, \cdots, n\}$, we have $|a_{k}| \leqslant \lambda(n) \cdot \max \left\{|a_{1}|,|a_{n}|\right\}$. (2009 China Western Mathematical Olympiad)
53. $\lambda(n)_{m n}=\frac{n+1}{n-1}$.

First, take $a_{1}=1, a_{2}=-\frac{n+1}{n-1}, a_{k}=-\frac{n+1}{n-1}+\frac{2 n(k-2)}{2(n-1)(n-2)}, k=3,4, \cdots$,
n. Then it satisfies $a_{1}+a_{2}+\cdots+a_{n}=0$ and $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots, n-1$. At this point, $\lambda(n) \geqslant \frac{n+1}{n-1}$

Next, we prove that when $\lambda(n)=\frac{n+1}{n-1}$, for all $k \in\{1,2, \cdots, n\}$, we have $\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \{\mid$ $\left.a_{1}|,| a_{n} \mid\right\}$

Since $2 a_{k} \leqslant a_{k-1}+a_{k+1}$, we have $a_{k+1}-a_{k} \geqslant a_{k}-a_{k-1}$, thus, $a_{n}-a_{n-1} \geqslant a_{n-1}-$ $a_{n-2}^{-} \geqslant \cdots \geqslant a_{2}-a_{1}$, so
$$\begin{aligned}
(k-1)\left(a_{n}-a_{1}\right)= & (k-1)\left[\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right] \geqslant \\
& (n-1)\left[\left(a_{k}-a_{k-1}\right)+\left(a_{k-1}-a_{k-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right]= \\
& (n-1)\left(a_{k}-a_{1}\right)
\end{aligned}$$

Therefore,
$$a_{k} \leqslant \frac{k-1}{n-1}\left(a_{n}-a_{1}\right)+a_{1}=\frac{1}{n-1}\left[(k-1) a_{n}+(n-k) a_{1}\right]$$

Similarly from equation (1), for a fixed $k$, and $k \neq 1, n$, when $1 \leqslant j \leqslant k$, $a_{j} \leqslant \frac{1}{k-1}[(j-1)$ $\left.a_{k}+(k-j) a_{1}\right]$.

When $k \leqslant j \leqslant n$, $a_{j} \leqslant \frac{1}{n-k}\left[(j-k) a_{n}+(n-j) a_{k}\right]$, so
$$\begin{array}{c}
\sum_{j=1}^{k} a_{j} \leqslant \frac{1}{k-1} \sum_{j=1}^{k}\left[(j-1) a_{k}+(k-j) a_{1}\right]=\frac{k}{2}\left(a_{1}+a_{k}\right) \\
\sum_{j=k}^{n} a_{j} \leqslant \frac{1}{n-k} \sum_{j=k}^{n}\left[(j-k) a_{n}+(n-j) a_{k}\right]=\frac{n+1-k}{2}\left(a_{k}+a_{n}\right)
\end{array}$$

Adding them together, we get
$$\begin{aligned}
a_{k}= & \sum_{j=1}^{k_{j}} a_{j}+\sum_{j=k}^{n} a_{j} \leqslant \frac{k}{2}\left(a_{1}+a_{k}\right)+\frac{n+1-k}{2}\left(a_{k}+a_{n}\right)= \\
& -\frac{k}{2} a_{1}+\frac{n+1}{2} a_{k}+\frac{n+1-k}{2} a_{n}
\end{aligned}$$

Therefore,
$$a_{k} \geqslant-\frac{1}{n-1}\left[k a_{1}+(n+1-k) a_{n}\right]$$

From equations (1) and (2), we get $\left|a_{k}\right| \leqslant \max \frac{1}{n-1}\left|(k-1) a_{n}+(n-k) a_{1}\right|, \left.\frac{1}{n-1} \right\rvert\, k a_{1}+$ $\left.(n+1-k) a_{n} \mid\right\} \leqslant \frac{n+1}{n-1} \max \left\{\left|a_{1}+,\right| a_{n} \mid\right\}, k=2,3, \cdots, n-1$

In conclusion, $\lambda(n)_{m n}=\frac{n+1}{n-1}$.