Is there a $t$ such that this function is positive? Let $b>0$ and $p (x,y) := \frac{1}{\sqrt{2\pi}}\exp (- \frac{(x-y)^2}{2})$ denote the Gaussian kernel. Set $1- \alpha := \int_{-b}^{b} p(0,y) \text d y$.
Define the function $\psi : [0,b]\times[0, b] \to \Bbb R$ by
$$\psi (t , c):= \int_{c}^\infty p(0,y)\text d y - \frac{1}{1-\alpha} \int_{-b}^{b} p(0,y) \int_{t+c}^{\infty} p(y,x)\text d x \text dy
$$
Let $Y, X \sim \mathcal{N} (0,1)$ standardnormal and independent. We can rewrite $\psi$ as follows.
$$\psi (t,c) = \Bbb P (Y \geq c) - \Bbb P (X+Y \geq t+c | \vert X\vert \leq b)$$
If we fix $c$ we observe that $t\mapsto \psi (t,c)$ is strictly increasing and one can compute (this comes from black boxes) that $\psi(0,c)\leq 0$ and $\psi (b,c)\geq 0$. Due to continuity there must be an unique $t_c \in [0,b]$ with $\psi (t_c ,c ) = 0$.
Additionally, from the same black box I know that $t_{b} < b$.
What I want to know is whether it is true that
$$\sup_{c\in [0,b]} t_c = t_b$$
or equivalently
$$\psi(t_b ,c ) \geq 0 \quad \forall c\in [0,b]$$
1) My first attempt: Observing that $\psi (t_b , 0) = \frac 1 2 - \Bbb P (X+Y \geq t_b| \vert X\vert \leq b) \geq 0$ gave me the idea to try whether
$$c\mapsto \psi (t_b ,c ) \text{ or more generally } c\mapsto \psi (t,c)$$
is decreasing, but could not figure it out.
A computation in python suggests that this is true for $t_b$ on $c\in[0,b]$ (but not for all $c$) and not true for some $t< t_b$.
2) Second idea: By intuition I guessed that $\Bbb P (X+Y \geq t+c | \vert X\vert \leq b) \leq \Bbb P (X+Y \geq t+c) $. This is indeed true and so one could try to show that
$$\Bbb P (Y \geq c) - \Bbb P (X+Y \geq t_b+c) \geq 0 \quad\forall c\in[0,b]$$
but I think that this is just not the case.
 A: I think I have found a way to solve this by ansatz 1. The property $c\mapsto \psi (t_b , c)$ decreasing is equivalent to
$$0 \geq \frac{\partial}{\partial c} \psi (t_b , c) = f(t_b + c) - g(c)\quad \forall c\in[0,b]$$
where $f(x) := \frac{1}{1-\alpha} \int_{-b}^{b} p(0,y)p(y,x) \text d y$ and $g(x) := p(0,x)$. Note that both $f$ and $g$ are symmetric densities with $f(0) < g(0)$. Since both integrate to $1$ they must have at least one intersection point on the positive real line (and thus the same on the negatives). Let $x_0$ denote the intersection point in $(-\infty , 0)$.
Let $b\ge t>0$ and define $f_t (x) := f(x+t)$. Since both $f$ and $g$ are radially decreasing we have that $f_t (0) <f(0)<g(0)$. Furthermore, we have that $f_t(x_0 - t)=f(x_0) =g(x_0) > g(x_0 -t)$. Due to continuity there has to be a intersection point of $f_t$ and $g$ in $(-\infty , 0)$.
Claim: $f_t$ and $g$ have at most two intersection points.
Given the claim we can proceed as follows: Suppose the remaining intersection point is contained in $[0,b]$. The claim yields that then $f_t (x) > g(x)$ for all $x> b$. By definition of $t_b$ we have that
$$0 = \psi (t_b , b) =\int_b^\infty ( g(x) -f_{t_b}(x) ) \text d x < 0$$
which is a contradiction. Therefore an intersection point in $[0,\infty)$ can only occur in $[b, \infty)$. By continuity and the fact that $f_{t_b} (0) < g(0)$ this means that $f(t_b + c) \le g(c)$ for all $c\in [0,b]$, which yields the statement.
Proof of the claim:
$x$ is an intersection point if and only if $\phi (x) := \log ( \frac{f_t (x)}{g(x)} ) = 0$. Now consider
$$ \phi ' (x) =  \frac{f '(x +t) }{f (x+t)} - \frac{g'(x)}{g(x)} = \frac{\int_{-b}^{b} p(0,y) (y - (x+t))p(y,x+t) \text d y}{\int_{-b}^{b} p(0,y) p(y,x+t) \text d y} +x  = \frac{\int_{-b}^{b}(y - t) p(0,y) p(y,x+t) \text d y}{\int_{-b}^{b} p(0,y) p(y,x+t) \text d y}$$
For every $x$ this can only be equal to $0$ if
$$0 = \int_{-b}^{b}(y - t) p(0,y) p(y,x+t) \text d y = \int_0^{b-t} z p(z+t,0)p(z,x)\text d z - \int_0^{b+t} z p(z-t,0) p(-z,x) \text d z$$
which can only be the case if
$$\varphi (x) := \frac{\int_0^{b-t} z p(z+t,0)p(z,x)\text d z}{\int_0^{b+t} z p(z-t,0) p(-z,x) \text d z} = 1$$
Claim: $\varphi$ is strictly increasing
For this note that it suffices to show that for all $z_0 \in (0,b-t), z_1 \in (0,b+t)$ we have that
$$x \mapsto \frac{z_0 p (0,t+z_0)p(z_0,x)}{z_1 p(0,z_1-t)p(-z_1,x)}$$
is strictly increasing. Transforming this yields that this mapping is
$$x\mapsto \frac{z_0 p (0,t+z_0)}{z_1 p(0,z_1-t)} e^{-\frac{1}{2} (z_0^2 -z_1^2)} e^{(z_0+z_1)x}$$
which is indeed strictly increasing in $x$. Looking back this means that
$\phi ' (x) = 0$ occurs for at most one $x$. This means that $\phi $ has at most one local extremum and cannot be constant anywhere. In view of the mean value theorem this means that $\phi$ can have at most two roots.
