For the $H^1 \subset L^2 \subset H^{-1}$ case.

Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.

Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon' \rangle \\ &= -\langle v,f'(u) u_\epsilon' \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.

Now for the Gelfand triple case, I'm not sure what to do. I needed to get some smoothing in the above argument.

For the $H^1 \subset L^2 \subset H^{-1}$ case.

Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.

Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon' \rangle \\ &= -\langle v,f'(u) u_\epsilon' \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.

For the $H^1 \subset L^2 \subset H^{-1}$ case.

Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.

Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon \rangle \\ &= -\langle v,f'(u) u_\epsilon \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.

Now for the Gelfand triple case, I'm not sure what to do. I needed to get some smoothing in the above argument.

For the $H^1 \subset L^2 \subset H^{-1}$ case.

Let $K_\epsilon$ be a mollifier then for $u\in L^2(0,T;L^2)$ we have that $u_\epsilon = K_\epsilon\ast u$ is actually smooth. In fact, $\lim_{\epsilon \rightarrow 0} u_\epsilon$ converges to $u$ in $L^p_{loc}$ for all $p\in [1,\infty)$. Since $f$ is differentiable, then it is continuous. So, we have the $\lim_{\epsilon \rightarrow 0} f(u_\epsilon) = f(u)$ in the same spaces as before.

Consider the following, \begin{align*} \langle v',f(u_\epsilon)\rangle &= -\langle v, f'(u_\epsilon) u_\epsilon' \rangle \\ &= -\langle v,f'(u) u_\epsilon' \rangle + \langle v,\left[f'(u) - f'(u_\epsilon)\right] u_\epsilon'\rangle \end{align*} A useful fact to use is that since $f$ has bounded derivative then $f$ is Lipshitz. Using this fact and taking $\epsilon$ to zero gives you what you want.

Now for the Gelfand triple case, I'm not sure what to do. I needed to get some smoothing in the above argument.