LM317 lower voltage limiting
Assuming you want to set the minimum/maximum voltage range of the external potentiometer, you just use two extra resistors. If you mean something else you'll need to clarify what it is you really want.







simulate this circuit &ndash; Schematic created using CircuitLab

The LM317 will change \$V_{out}\$ such that \$V_{adj} = 1.25V\$. This configuration has the relationship (ignoring adjust pin current)
\begin{gather}
V_{out} = \frac{V_{adj}}{R_{1}^{-} + R_3} (R_2 + R_1 + R_3)
\end{gather}
where \$R_{1}^-\$ denotes the "lower" portion of the potentiometer. At the maximum stop, \$R_{1}^- = 0\$, so
\begin{gather}
V_{out,max} = \frac{V_{adj}}{R_3} (R_2 + R_1 + R_3)
\end{gather}
At the minimum stop, \$R_{1}^- = R_1\$, so
\begin{gather}
V_{out,min} = \frac{V_{adj}}{R_3+R_1} (R_2 + R_1 + R_3)
\end{gather}
This gives 2 equations with three unknowns, and the easiest way to solve this is to pick an arbitrary value for one of the resistors. For example, say you fix \$R_1\$. Then you get (assuming my algebra is correct)
\begin{gather}
R_3 = \frac{R_1 V_{out,min}}{V_{out,max}-V_{out,min}}\\
R_2 = \left(\frac{V_{out,max}}{V_{adj}}-1\right) R_3 -R_1
\end{gather}
As long as \$R_1\$ isn't too big (say, less than 10k) or you have some leeway in the min/max limits, you can ignore the adjust pin current (as I've done for my analysis). If \$R_1\$ is very big or you need very precise bounds (not sure why you would need this), then you'll need to account for it in calculating the resistor values but the same design should work.

Note that I'm glossing over a few details of how to properly use the LM317 (for example, I'm not including decoupling caps). See the LM317's datasheet for more information.