7. Let the function $f(x)$ have a derivative $f^{\prime}(x)$ on $\mathbf{R}$, for any $x \in \mathbf{R}$ we have $f(x)+f(-x)=x^{2}$, and on $(0,+\infty)$, $f^{\prime}(x)>x$. If $f(1+a)-f(1-a) \geqslant 2 a$, then the range of the real number $a$ is $\qquad$.
Solve:
$$
f(x)-\frac{x^{2}}{2}+f(-x)-\frac{(-x)^{2}}{2}=0 \Rightarrow g(x)=f(x)-\frac{x^{2}}{2} \text{ is an odd function, and on } (0,+\infty) \text{, } g^{\prime}(x)=f^{\prime}(x)-x>0, \text{ thus } g(x) \text{ is an increasing function on } \mathbf{R}.
$$
$$
\begin{array}{l}
f(1+a)-f(1-a) \geqslant 2 a=\frac{(1+a)^{2}-(1-a)^{2}}{2} \\
\Rightarrow f(1+a)-\frac{(1+a)^{2}}{2} \geqslant f(1-a)-\frac{(1-a)^{2}}{2} \Rightarrow g(1+a) \geqslant g(1-a) \Rightarrow a \geqslant 0 .
\end{array}
$$

Therefore, the range of the real number $a$ is $[0,+\infty)$.