An example of a Cameron–Martin space I am trying to compute an example for a Cameron–Martin space, following an exercise in M. Hairer's notes on SPDEs.
The problem is the following: consider $\mathcal{C}[0,1]$ endowed with Wiener measure. Prove that the Cameron Martin space (or "reproducing kernel" space) associated to this measure is $W^{1,1}.$
To see that an element in the range of the covariance operator has one weak derivative is not difficult. But my computations actually show that the weak derivative is bounded, not only square integrable.
In fact let $\mu \in \mathcal{M}[0,1] = C[0,1]^*,$ then $$Q(\mu) (s)= \int_0^1s \wedge u \text{ } \mu(du) = \int_0^s\int_u^1 \mu(dv) du.$$ Now if we look at our weak derivative $\frac{d}{ds} Q(\mu) (s)= \int_s^1 \mu(dv)$ we see that this function is actually bounded, since $\mu$ is of bounded total variation. 
So the question is: what did I do wrong? Where have I missed the $L^2$ norm of the derivative? 
 A: I actually solved the issue, and as always these "easy" exercises show that you need to properly understand the definitions.
The Cameron–Martin space $\mathcal{H}$ is the completion of the range of the covariance operator, which we call $\overset{\circ}{\mathcal{H}}$, endowed with the scalar product induced by $Q$ (hence it is in particular a Hilbert space).
What I did not do above is see what the scalar product induced by $Q$ is. So let $a, b \in \overset{\circ}{\mathcal{H}}, a = Q \mu , b = Q \nu.$ We get, by abusing the notiation of scalar product (note that we have Banach spaces so in general the written product is a duality pairing): $$<a,b>_{\overset{\circ}{\mathcal{H}}} = <Q \mu , \nu> = \int_0^1 Q \mu (s) \: \nu (ds) = \int_0^1 \left( \int_0^s \left( \int_u^1 \mu(dv) \right) \: du \right) \: \nu(ds) = $$ $$ = \int_0^1 \left(\int_u^1 \mu(dv) \right) \left( \int_u^1 \nu(ds)\right) \: du = <a,b>_{W^{1,2}_0} $$
So we get the $W^{1,2}_0$ norm, where by $W^{1,2}_0$ we refer to the Sobolev space with Dirichlet condition in zero (otherwise we would not really have a norm).
This means that if we take the completion of the space w.r.t this norm, we actually find $W^{1,2}_0 = \mathcal{H}.$
