11. (15 points) Let $f(x)$ be a function defined on the domain $D$. If for any real number $\alpha \in(0,1)$ and any two numbers $x_{1}, x_{2}$ in $D$, it always holds that
$$
f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \leqslant \alpha f\left(x_{1}\right)+(1-\alpha) f\left(x_{2}\right),
$$
then $f(x)$ is called a $C$ function defined on $D$.
(1) Given that $f(x)$ is a $C$ function on $\mathbf{R}$, and $m$ is a given positive integer. Let $a_{n}=f(n)(n=0,1, \cdots, m)$, and $a_{0}=0, a_{m}=2 m$, and denote $S_{f}=a_{1}+a_{2}+\cdots+a_{m}$. For any function $f(x)$ that satisfies the conditions, try to find the maximum value of $S_{f}$.
(2) If $f(x)$ is a function defined on $\mathbf{R}$ with the smallest positive period $T$, prove that $f(x)$ is not a $C$ function on $\mathbf{R}$.
11. (1) For any $n(0 \leqslant n \leqslant m)$, take
$$
x_{1}=m, x_{2}=0, \alpha=\frac{n}{m} \in[0,1] \text {. }
$$

Since $f(x)$ is a $C$ function on $\mathbf{R}$, $a_{n}=f(n)$, and $a_{0}=0, a_{m}=2 m$, we have
$$
\begin{array}{l}
a_{n}=f(n)=f\left(\alpha x_{1}+(1-\alpha) x_{2}\right) \\
\leqslant \alpha f\left(x_{1}\right)+(1-\alpha) f\left(x_{2}\right)=\frac{n}{m} \times 2 m=2 n .
\end{array}
$$

Thus, $S_{f}=a_{1}+a_{2}+\cdots+a_{m}$
$$
\leqslant 2(1+2+\cdots+m)=m^{2}+m \text {. }
$$

We can prove that $f(x)=2 x$ is a $C$ function, and it makes $a_{n}=$ $2 n(n=0,1, \cdots, m)$ all hold. In this case,
$$
S_{f}=m^{2}+m \text {. }
$$

In summary, the maximum value of $S_{f}$ is $m^{2}+m$.
(2) Assume $f(x)$ is a $C$ function on $\mathbf{R}$.
If there exists $mf(n)$, let $x_{1}=n, x_{2}=n-T$, $\alpha=1-\frac{n-m}{T}$, a contradiction can also be derived.
Therefore, $f(x)$ is a constant function on $[0, T)$.
Since $f(x)$ is a periodic function with period $T$, $f(x)$ is a constant function on $\mathbf{R}$, which contradicts the fact that the smallest positive period of $f(x)$ is $T$.
Hence, $f(x)$ is not a $C$ function on $\mathbf{R}$.