Let $H=$ the collection of all absolutely continuous functions, $ f:[0,1]\to \mathbb{C}$, such that $f(0)=0$ and $f'\in L^2(0,1)$. If $<f,g>=\int^1_0f'(t)\overline{g'(t)}$ for $f$ and $g$ in $H$, fix $t$,$0<t\leq 1$ define $L:H\to \mathbb{C}$ by $L(h)=h(t)$,find $\|L\|$ and $h_0$ such that $L(h)=<h,h_0>$ for every $h$ in $H$.
it is easy to know $H$ is a Hilbert space, but I do not know to how to use condition absolutely continuous to find $\|L\|$.
The space $H$ that you are defining is actually a subspace of the homogeneous Sobolev $\dot H^1([0, 1])$. In 1D, $f \in \dot H^1([0, 1])$ implies $f$ has to be absolutely continuous.
Edit: Fix $t \in (0, 1]$ and define $L_t(h) = h(t)= \int^t_0 h'(\tau)\ d\tau$ (Here we used the fact $h$ is absolutely continuous). Moreover, it follows \begin{align} |L_t(h)| = \left|\int^t_0h'(\tau)\ d\tau\right| = \left|\int^1_0 \chi_{[0, t]}(\tau)h'(\tau)\ d\tau\right| \leq \sqrt{t}|| h||_H \end{align} which means \begin{align} ||L_t||_{H^\ast} \leq \sqrt{t}. \end{align} Consider a function $h$ defined by \begin{align} h(x) = \begin{cases} \sqrt{t}& \text{ if } x \geq t\\ \frac{x}{\sqrt{t}} & \text{ if } x<t \end{cases}. \end{align} Let us check $h \in H$ and $|| h||_H=1$. Observe \begin{align} h'(x) = \begin{cases} 0 & \text{ if } x \geq t \\ \frac{1}{\sqrt{t}} & \text{ if } x<t \end{cases} \end{align} which means \begin{align} \int^1_0 |h'(x)|^2\ dx = \frac{1}{t}\int^t_0 dx = 1. \end{align} Lastly, observe \begin{align} |L_t(h)| = \left|\int^t_0 \frac{1}{\sqrt{t}} d\tau \right| = \sqrt{t}. \end{align} Thus, it follows $||L_t||_{H^\ast} = \sqrt{t}$.
To spell out explicitly what is $h_0$ (more or less constructed already in the other answer): Using
$$L_t(h) = h(t)= \int^t_0 h'(\tau)\ d\tau = \int_0^1 h' (\tau) \chi_{[0,t]}(\tau) d\tau, $$ where $\chi_Y$ is the characteristic function of the set $Y\subset [0,1]$. From the definition of the inner product, you want $h_0$ so that $h_0' = \chi_{[0,t]}$. Then one can choose
$$h_0(\tau) = \begin{cases} \tau & \text{ if }\tau \in [0,t], \\ t & \text{ if } \tau \in [t, 1]\end{cases}$$
Then $$h'_0(\tau) = \begin{cases} 1 & \text{ if }\tau \in [0,t), \\ 0 & \text{ if } \tau \in (t, 1]\end{cases}.$$
Thus $h_0' = \chi_{[0,t]}$ and thus
$$ \langle h, h_0\rangle = \int_0^1 h' h'_0 = \int_0^1 h' \chi_{[0,t]} = L_t(h).$$
