Using the definition for the Fourier transform, prove that this operation is linear.
To prove that the Fourier transform is a linear operation, we need to show that it satisfies the properties of additivity and scalar multiplication.
Let's start with the definition of the Fourier transform. For a given function \(f(t)\), the Fourier transform is defined as:
\(F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt\)
Now, let's consider two functions \(f_1(t)\) and \(f_2(t)\) and their Fourier transforms \(F_1(\omega)\) and \(F_2(\omega)\) respectively.
1. Additivity:
To show additivity, we need to prove that \(F(f_1(t) + f_2(t)) = F_1(\omega) + F_2(\omega)\).
Using the definition of the Fourier transform, we have:
\(F(f_1(t) + f_2(t)) = \int_{-\infty}^{\infty} (f_1(t) + f_2(t))e^{-i\omega t} dt\)
Expanding the integral, we get:
\(F(f_1(t) + f_2(t)) = \int_{-\infty}^{\infty} f_1(t) e^{-i\omega t} dt + \int_{-\infty}^{\infty} f_2(t) e^{-i\omega t} dt\)
This can be written as:
\(F(f_1(t) + f_2(t)) = F_1(\omega) + F_2(\omega)\)
Hence, the Fourier transform satisfies the property of additivity.
2. Scalar Multiplication:
To prove scalar multiplication, we need to show that \(F(kf(t)) = kF(\omega)\), where \(k\) is a scalar.
Using the definition of the Fourier transform, we have:
\(F(kf(t)) = \int_{-\infty}^{\infty} kf(t) e^{-i\omega t} dt\)
Rearranging the integral and applying the scalar multiplication, we get:
\(F(kf(t)) = k\int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt\)
Since \(k\) is a constant, we can pull it out of the integral, giving:
\(F(kf(t)) = kF(\omega)\)
Thus, the Fourier transform satisfies the property of scalar multiplication.
Therefore, since the Fourier transform satisfies the properties of additivity and scalar multiplication, we have proven that it is a linear operation.