So, this should be really simple: Let $$\varphi(x):=e^{-(cx)^2}\;\;\;\text{for }x\in\mathbb R$$ for some $c>0$ and $$\varphi_x(y):=\varphi\left(|x-y|\right)\;\;\;\text{for }x,y\in \mathbb R.$$ How should I use $c$ if I want that $$\varphi_x(x\pm h)<\varepsilon\;\;\;\text{for all }x\in\mathbb R\tag1,$$ where $h>0$ and $\varepsilon$ are given?
We should easily see that $$\varphi_x(y)<\varepsilon\Leftrightarrow|x-y|^2>\frac{-\ln\varepsilon}{c^2}\tag2$$ and hence $$\varphi_x(x\pm h)<\varepsilon\Leftrightarrow h^2>-\frac{\ln\varepsilon}{c^2}\tag3.$$
Now, assume I want to place the kernel $\varphi$ at $0.5$ such that the set where it is $>\varepsilon$ is precisely $[0,1)$. This amounts to taking $x=0.5$ and $h=\frac12$. However, if I choose $c$ accordingly and take $\varepsilon=1e-10$ for example, then this is what I get (https://www.wolframalpha.com/input?i2d=true&i=plot+exp%5C%2840%29-%5C%2840%29-ln%5C%2840%291e-10%5C%2841%29*4%5C%2841%29*Power%5B%5C%2840%290.5-y%5C%2841%29%2C2%5D%5C%2841%29+from+y%3D0+to+1)
This is not what I intended; the "relevant support" of the function should have length $1$. If I now replace $h$ by $h=1$, then I get
what is what I was looking for in the first place. However, this plot contradicts my derivation above. What am I missing here?
According to your (correct) derivation, since in your 1st plot $h=0.5$ and $\varepsilon=10^{-10},$ you rightly took $$c^2=\frac{-\ln\varepsilon}{h^2}=40\ln(10).$$ Your plot is not surprising: the "relevant support" (i.e. the interval where $\varphi>10^{-10}$), is really $(0,1),$ but your $\varepsilon$ is so tiny that you allow values of $\varphi$ indiscernible from $0$ on this support.
And for the second one ($h=1$), keeping $\varepsilon=10^{-10},$ you rightly took $$c^2=\frac{-\ln\varepsilon}{h^2}=10\ln(10).$$ But your plot does not show the full "relevant support", wich is now (-0.5,1.5) and has the same problem as the first one.