26. Let $f(x)=\left\{\begin{array}{l}1, x \text { is rational } \\ 0, x \text { is irrational }\end{array}\right.$ For all $x$, the function $g(x)$ that satisfies $x f(x) \leqslant g(x)$ is $(\quad$.
(A) $g(x)=\sin x$
(B) $g(x)=x$
(C) $g(x)=x^{2}$
(D) $g(x)=|x|$
(E) *
26. (1).
$$
\begin{array}{l}
\because \sin 2<2=2 f(2),(-\sqrt{2})<0=(-\sqrt{2}) \cdot f(-\sqrt{2}), \\
\left(\frac{1}{2}\right)^{2}<\frac{1}{2}=\frac{1}{2} f\left(\frac{1}{2}\right) .
\end{array}
$$
$\therefore(A) 、(B) 、(C)$ do not meet the requirements of the problem.
Also, $x f(x)=\left\{\begin{array}{l}x, x \text { is a rational number. } \\ 0, x \text { is an irrational number. }\end{array}\right.$
It is easy to see that $x \leqslant|x|$.