Final answer:
L_c, defined as L_c{f(t)}(s) = ∫(c to ∞) e^(-st) f(t) dt, is a linear transformation because it satisfies both the additivity and homogeneity conditions. This confirmation was done by demonstrating that L_c{f(t) + g(t)} = L_c{f(t)} + L_c{g(t)} and L_c{αf(t)} = αL_c{f(t)} for arbitrary functions f(t), g(t) and scalar α.
Step-by-step explanation:
The question asks if L_c defined by the integral L_c{f(t)}(s) = ∫(c to ∞) e^(-st) f(t) dt is a linear transformation. A linear transformation must satisfy two conditions: additivity and homogeneity. To prove that L_c is a linear transformation, we consider two arbitrary functions f(t) and g(t) and a scalar α.
Additivity tells us that L_c{f(t) + g(t)} should be equal to L_c{f(t)} + L_c{g(t)}. Let's check this:
L_c{f(t) + g(t)} = ∫(c to ∞) e^(-st) (f(t) + g(t)) dt = ∫(c to ∞) e^(-st) f(t) dt + ∫(c to ∞) e^(-st) g(t) dt = L_c{f(t)} + L_c{g(t)}.
Homogeneity requires that L_c{αf(t)} equals αL_c{f(t)}. Let's verify:
L_c{αf(t)} = ∫(c to ∞) e^(-st) (αf(t)) dt = α ∫(c to ∞) e^(-st) f(t) dt = αL_c{f(t)}.
Therefore, since both additivity and homogeneity are fulfilled, L_c is indeed a linear transformation.