It is well known that an abstract Wiener space can be constructed for the Brownian bridge pinned to 0 at both $t = 0$ and $t = T$: The sample space is the loop space of all continuous paths which start and end at 0, $C_{0,0}[0,T]$, and the Hilbert space $\mathcal{H}$ is its subset of absolutely continuous paths with square-integrable first derivatives under the inner product $$\left(f, g\right)_\mathcal{H} = \int_0^T \left(\dot{f} + \frac{f}{T-\tau}\right) \left(\dot{g} + \frac{g}{T-\tau}\right) \mathrm{d}\tau.$$ The covariance function is $$\mathrm{Cov}(B_s, B_t) =: a(s,t) = \min(s, t) - \frac{st}{T},$$ which is easily seen to be an element of $\mathcal{H}$ and is thus a valid reproducing kernel, as required. Note that $a(s, T) = 0$ (as required by the pinning). So far, so good.

However, once we pin the bridge to a non-zero final value $b = B_T$, something apparently has to change. The covariance must, of course, be the same regardless of the pinned value, but the sample space and Cameron–Martin Hilbert space are not, since they must now be pinned to $b \neq 0$. Hence the covariance function is no longer a valid reproducing kernel because it still ends at $a(s, T) = 0$, regardless of $b$, and is therefore not an element of $\mathcal{H}$ any more!

Edit: I now noticed that, of course, if the final value is pinned to $b \neq 0$, then the functions no longer form a vector space, so the question is actually much more fundamental: Is it even possible to define an abstract Wiener space in this case? If so, how?

Hence my question is: What do I need to modify to properly accommodate a non-zero pinned final value?

Since the covariance is a fundamental property of the underlying stochastic process, it seems I can't actually change anything about that, but this then implies that it will never be a reproducing kernel for the Cameron–Martin space, which seems to be a contradiction.

In fact, more generally, this must be a problem for any process that is pinned to a non-zero final value.

I thought I might be able to fudge my way out of this by applying the Cameron–Martin theorem with a simple linear drift $h(\tau) = k\tau$ with $k = b/T$ to turn the 0-pinned case into a $b$-pinned one, but this $h(\tau)$ turns out to have infinite norm $(h, h)_\mathcal{H}$ for $b \neq 0$ (which isn't surprising considering that it is not an element of the original Cameron–Martin space!).

In introductions to abstract Wiener spaces, the sample paths usually form a Banach space; so, in particular, the sum of two sample paths is a valid sample path and also an element of the Banach space. Then the Gaussian measure is constructed as a measure on this Banach space, the Cameron–Martin space is a subspace thereof, etc.

However, when the stochastic process is pinned to some non-zero value, then the addition of two sample paths no longer yields a valid sample path, hence the sample paths do not form a Banach space and the whole construction breaks down—so how are abstract Wiener spaces constructed for such pinned processes?

A simple example of this is a Brownian bridge, which is typically constructed to be pinned to zero at both time 0 and some time $T > 0$, but what if it is pinned to some value $b \neq 0$ at time $T$? Is the only way out to take the zero-pinned Wiener space and add $b\tau/T$ to every sample path? So then in a sense there is no true Wiener space for non-zero-pinned processes (since this additional term takes it out of the Banach space)? Also, how do I know that a linear drift is correct instead of some other drift function?

Note: I substantially modified this question after realizing that I had originally kind of missed the point.

It is well known that an abstract Wiener space can be constructed for the Brownian bridge pinned to 0 at both $t = 0$ and $t = T$: The sample space is the loop space of all continuous paths which start and end at 0, $C_{0,0}[0,T]$, and the Hilbert space $\mathcal{H}$ is its subset of absolutely continuous paths with square-integrable first derivatives under the inner product $$\left(f, g\right)_\mathcal{H} = \int_0^T \left(\dot{f} + \frac{f}{T-\tau}\right) \left(\dot{g} + \frac{g}{T-\tau}\right) \mathrm{d}\tau.$$ The covariance function is $$\mathrm{Cov}(B_s, B_t) =: a(s,t) = \min(s, t) - \frac{st}{T},$$ which is easily seen to be an element of $\mathcal{H}$ and is thus a valid reproducing kernel, as required. Note that $a(s, T) = 0$ (as required by the pinning). So far, so good.

However, once we pin the bridge to a non-zero final value $b = B_T$, something apparently has to change. The covariance must, of course, be the same regardless of the pinned value, but the sample space and Cameron–Martin Hilbert space are not, since they must now be pinned to $b \neq 0$. Hence the covariance function is no longer a valid reproducing kernel because it still ends at $a(s, T) = 0$, regardless of $b$, and is therefore not an element of $\mathcal{H}$ any more!

Hence my question is: What do I need to modify to properly accommodate a non-zero pinned final value?

Since the covariance is a fundamental property of the underlying stochastic process, it seems I can't actually change anything about that, but this then implies that it will never be a reproducing kernel for the Cameron–Martin space, which seems to be a contradiction.

In fact, more generally, this must be a problem for any process that is pinned to a non-zero final value.

I thought I might be able to fudge my way out of this by applying the Cameron–Martin theorem with a simple linear drift $h(\tau) = k\tau$ with $k = b/T$ to turn the 0-pinned case into a $b$-pinned one, but this $h(\tau)$ turns out to have infinite norm $(h, h)_\mathcal{H}$ for $b \neq 0$ (which isn't surprising considering that it is not an element of the original Cameron–Martin space!).

It is well known that an abstract Wiener space can be constructed for the Brownian bridge pinned to 0 at both $t = 0$ and $t = T$: The sample space is the loop space of all continuous paths which start and end at 0, $C_{0,0}[0,T]$, and the Hilbert space $\mathcal{H}$ is its subset of absolutely continuous paths with square-integrable first derivatives under the inner product $$\left(f, g\right)_\mathcal{H} = \int_0^T \left(\dot{f} + \frac{f}{T-\tau}\right) \left(\dot{g} + \frac{g}{T-\tau}\right) \mathrm{d}\tau.$$ The covariance function is $$\mathrm{Cov}(B_s, B_t) =: a(s,t) = \min(s, t) - \frac{st}{T},$$ which is easily seen to be an element of $\mathcal{H}$ and is thus a valid reproducing kernel, as required. Note that $a(s, T) = 0$ (as required by the pinning). So far, so good.

However, once we pin the bridge to a non-zero final value $b = B_T$, something apparently has to change. The covariance must, of course, be the same regardless of the pinned value, but the sample space and Cameron–Martin Hilbert space are not, since they must now be pinned to $b \neq 0$. Hence the covariance function is no longer a valid reproducing kernel because it still ends at $a(s, T) = 0$, regardless of $b$, and is therefore not an element of $\mathcal{H}$ any more!

Edit: I now noticed that, of course, if the final value is pinned to $b \neq 0$, then the functions no longer form a vector space, so the question is actually much more fundamental: Is it even possible to define an abstract Wiener space in this case? If so, how?

Hence my question is: What do I need to modify to properly accommodate a non-zero pinned final value?

Since the covariance is a fundamental property of the underlying stochastic process, it seems I can't actually change anything about that, but this then implies that it will never be a reproducing kernel for the Cameron–Martin space, which seems to be a contradiction.

In fact, more generally, this must be a problem for any process that is pinned to a non-zero final value.

I thought I might be able to fudge my way out of this by applying the Cameron–Martin theorem with a simple linear drift $h(\tau) = k\tau$ with $k = b/T$ to turn the 0-pinned case into a $b$-pinned one, but this $h(\tau)$ turns out to have infinite norm $(h, h)_\mathcal{H}$ for $b \neq 0$ (which isn't surprising considering that it is not an element of the original Cameron–Martin space!).