How to offset an input voltage of 0V ~ 10V to -400mv ~ +400mv?
Assuming your source output impedance is \$0\:\Omega\$ and the destination's input impedance is infinite and you have access to both \$\pm 10\:\textrm{V}\$ rails, then something like this would do it:







simulate this circuit &ndash; Schematic created using CircuitLab

Of course, I'm sure reality impinges and this won't help you for any number of practical reasons you didn't discuss about your circumstances.



Updated to include your \$\pm 5\:\textrm{V}\$ supplies:







simulate this circuit

If \$R_2=X=47\:\textrm{k}\Omega\$, then \$R_1\approx 56\:\textrm{k}\Omega\$, \$R_3\approx 294\:\textrm{k}\Omega\$, \$R_{TH_{in}}\approx 319\:\textrm{k}\Omega\$, and \$R_{TH_{out}}= 23.5\:\textrm{k}\Omega\$.



For those interested, here's a general solution approach (that may be adapted, as needed.) Suppose you know \$V_1\$, \$V_2\$, \$V_{I_{min}}\$, \$V_{I_{max}}\$, \$V_{O_{min}}\$, \$V_{O_{max}}\$, and the value for \$R\$ (which guarantees that the loading on the input will be more than that.) You can obviously then compute \$\Delta V_O = V_{O_{max}} - V_{O_{min}}\$ and \$\Delta V_I = V_{I_{max}} - V_{I_{min}}\$.

The schematic:







simulate this circuit

And the values for \$R_1\$ and \$R_2\$ are then:

$$\begin{align*}
R_1 &amp;= R\cdot\frac{\Delta V_O\cdot\left(V_2-V_1\right)}{V_2\cdot\left(\Delta V_I-\Delta V_O\right)-V_{O_{min}}\cdot V_{I_{max}}+V_{O_{max}}\cdot V_{I_{min}}} \\ \\
R_2 &amp;= R\cdot\frac{\Delta V_O\cdot\left(V_1-V_2\right)}{V_1\cdot\left(\Delta V_I-\Delta V_O\right)-V_{O_{min}}\cdot V_{I_{max}}+V_{O_{max}}\cdot V_{I_{min}}}
\end{align*}$$

Might be useful in a pinch where you know the input range and the output range. If the values aren't both positive then the circuit isn't realizable as shown. But the reason why will probably be obvious, too.