Can You Prove This Fourier Transform Property?

[dimensionless]

Here's my problem:

Suppose that f is an integrable function. If $$\lambda \ne 0$$ is a real number and $$g(x) = f(\lambda x + y)$$ proove that

$$G(k) = \frac{1}{\lambda} e^{ik/\lambda}F(\frac{k}{\lambda})$$

Where F and G are the Fourier transforms of f and g, respectively.

I've got the Fourier transform of f(x) as

$$F(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(x) e^{-ikx} dx$$

Likewise for g(x) I have

$$G(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty g(x) e^{-ikx} dx$$

for F(K/lambda) I have

$$F(K/\lambda) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(\lambda x + y) e^{-ik(\lambda x + y)/\lambda} d(\lambda x + y)$$

My guess is that the answer is relatively simple from here, but I'm not sure how to go about it. Does anyone understand this?

 

[nrqed]

Here's my problem:


I've got the Fourier transform of f(x) as

$$F(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(x) e^{-ikx} dx$$

Likewise for g(x) I have

$$G(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty g(x) e^{-ikx} dx$$

for F(K/lambda) I have

$$F(K/\lambda) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(\lambda x + y) e^{-ik(\lambda x + y)/\lambda} d(\lambda x + y)$$

You should simply apply the definition to g(x), so

$$G(k) = {1 \over {\sqrt{ 2 \pi}} } \int dx \,f(\lambda x + y) \, e^{-i kx}$$

Now, define a new variable x'= lambda x + y so that x=(x'-y)/lambda and dx= dx'/lambda.

You get

$$G(k) = { 1\over \lambda} {1 \over {\sqrt{ 2 \pi}}} \int \, dx' \, f(x') \, e^{-i k x'/ \lambda} e^{i k y / \lambda}$$

which directly gives the result you were looking for (I am asusming there was typo in the fomula you gave and that the y was missing in the exponent).

Patrick