14. Define $f(x)$ as an odd function on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$. If for any $x \in[a, a+2]$, $f(x+a) \geqslant 2 f(x)$, then the range of the real number $a$ is $\qquad$
14. $[\sqrt{2},+\infty)$.

From the problem, we know that for any real numbers $x, y$ satisfying the domain, $f(x y)=f(x) f(y)$.
Given that $f(x)$ is an odd function on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)$ is monotonically increasing. Therefore,
$$
\begin{array}{l}
f(x+a) \geqslant 2 f(x) \\
\geqslant f(\sqrt{2}) f(x)=f(\sqrt{2} x) \\
\Rightarrow x+a \geqslant \sqrt{2} x \Rightarrow x \leqslant \frac{a}{\sqrt{2}-1} .
\end{array}
$$

Since for $x \in[a, a+2]$, equation (1) always holds, we have,
$$
\frac{a}{\sqrt{2}-1} \geqslant a+2 \Rightarrow a \geqslant \sqrt{2} \text {. }
$$

Thus, $a \in[\sqrt{2},+\infty)$.