= Solution
{c}
Write
$$
\omega((v,f),(w,g))=f(w)-g(v)
$$
on $V=W\times W'$, where $W'$ is the <dual space>. The <dual pairing> is <linear> separately in the <vector> and the <linear functional>. For example, for <real numbers> $a,b$,
$$
\begin{aligned}
\omega(a(v_1,f_1)+b(v_2,f_2),(w,g))
&=(af_1+bf_2)(w)-g(av_1+bv_2)\\
&=a\omega((v_1,f_1),(w,g))+b\omega((v_2,f_2),(w,g)),\\
\omega((v,f),a(w_1,g_1)+b(w_2,g_2))
&=f(aw_1+bw_2)-(ag_1+bg_2)(v)\\
&=a\omega((v,f),(w_1,g_1))+b\omega((v,f),(w_2,g_2)).
\end{aligned}
$$
Thus $\omega$ is a <bilinear form>. It is also an <alternating bilinear form>, since
$$
\omega((v,f),(v,f))=0,\qquad
\omega((w,g),(v,f))=-\omega((v,f),(w,g)).
$$
To prove that this is a <nondegenerate bilinear form>, suppose $(v,f)$ pairs to zero with every $(w,g)$. Setting $g=0$ gives $f(w)=0$ for every $w\in W$, hence $f=0$. Setting $w=0$ gives $g(v)=0$ for every $g\in W'$. If $v\ne0$, <basis extension> supplies a <coordinate functional> with $g(v)=1$, a contradiction. Thus $v=0$. The <left radical> is zero, and the displayed change of sign implies the <right radical> is zero as well. Hence \b[$\omega$ is a <nondegenerate bilinear form>], making $V$ a <symplectic vector space>.
For the required example take the <coordinate subspace>
$$
U=W\times\{0\}.
$$
Using the definition of the <bilinear orthogonal complement> in part (b),
$$
\begin{aligned}
(w,g)\in U^\perp
&\ \Longleftrightarrow\ \omega((v,0),(w,g))=-g(v)=0\quad\text{for every }v\in W\\
&\ \Longleftrightarrow\ g=0.
\end{aligned}
$$
Therefore
$$
\boxed{U^\perp=W\times\{0\}=U.}
$$
This is a <Lagrangian subspace>. Its <dimension> is $\dim W$, exactly half of $\dim V=2\dim W$, so part (b)(i) is satisfied.
There is no contradiction with part (b)(ii): that part asks whether a <direct sum> is guaranteed, and the answer is already no. Moreover, when $W\ne0$, the present <alternating bilinear form> is not a <symmetric bilinear form>. Choose $v\ne0$ and $f\in W'$ with $f(v)=1$; then
$$
\omega((v,0),(0,f))=-1,\qquad \omega((0,f),(v,0))=1.
$$
It therefore does not satisfy the symmetry hypothesis of part (b)(ii). If $W=0$, then $V=U=0$, and both $U=U^\perp$ and $V=U\oplus U^\perp$ hold trivially.
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