= Solution
{c}
Put $e_h=u-u_h$ and $w=z-I_hz$. <Galerkin orthogonality> gives $a(e_h,I_hz)=0$, so the dual identity becomes
$$
\int_0^1e_h\psi=a(e_h,z)=a(e_h,w)=\int_0^1fw-a(u_h,w).
$$
On $I_i=(x_{i-1},x_i)$, $u_h'$ is constant and $w$ is zero at both endpoints. Consequently $\int_{I_i}u_h'w'=u_h'[w]_{x_{i-1}}^{x_i}=0$. Defining the element residual by
$$
\boxed{R(u_h)|_{I_i}=f-e^xu_h}
$$
gives
$$
\int_0^1(u-u_h)\psi=\sum_{i=1}^N\int_{I_i}R(u_h)(z-I_hz).
$$
No derivative-jump terms survive, because this <interpolation> error vanishes at every mesh node. This is the mechanism of the <explicit one-dimensional dual residual bound in L2>.
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