# Are the Fourier sine and cosine transforms injective?

Are the Fourier sine and cosine transforms defined by $$\mathcal{F}_s[f(x)](t)=\int_0^\infty f(x)\sin(x t)\text{d}x$$ and $$\mathcal{F}_c[f(x)](t)=\int_0^\infty f(x)\cos(x t)\text{d}x$$ injective? That is to say, does $$\mathcal{F_s[f(x)](t)}=\mathcal{F_s[g(x)](t)}$$ imply that $f(x)=g(x)$, and similarly for the Fourier cosine?

• Since the transforms are defined by integrals, the best you could possibly hope for is that $f=g$ almost everywhere. – Chris Eagle Jul 20 '13 at 8:24
• Do you know about the uniqueness theorem for the Fourier transform? Note that you can use that if you extend the function $f$ to $\mathbb{R}$. – AD. Jul 20 '13 at 8:30
• @ChrisEagle I am sure pbs is aware of that. No? – AD. Jul 20 '13 at 8:31

The whole story is a very long one but the executive summary is that the answer is yes, on interesting function spaces they are injective (and surjective, if you've defined the codomain in a good way). This is because the Fourier transform on $\mathbb{R}$ maps even functions to even functions, and maps odd functions to odd functions. Every function can be written as the sum of an even function and an odd function, so where the Fourier transform in injective, it is injective from even functions to even functions, and from odd functions to odd functions. But every even function is determined by it's values on the non-negative real line, and similarly for odd functions. Since $e^{ikx}=\cos(kx)+i \sin(kx)$, when $f(x)$ is even $$\int_{-\infty}^\infty e^{ikx} f(x)dx=\int_{-\infty}^\infty \cos(kx) f(x)dx=2\int_0^\infty \cos(kx) f(x)dx$$ and when $f(x)$ is odd $$\int_{-\infty}^\infty e^{ikx} f(x)dx=i\int_{-\infty}^\infty \sin(kx) f(x)dx=2i\int_0^\infty \sin(kx) f(x)dx.$$ Thus the Fourier cosine transform is the restriction to the non-negative reals of the Fourier transform of an even function, hence injective. Likewise for the Fourier sine transform and the Fourier transform of odd functions.
Now, as @ChrisEagle points out, functions zero except on null sets all get mapped to the zero function for reasonable definitions of the integral, which would be too bad except for the fact that these functions are considered negligible and are quotiented out e.g. in defining the $L^p$ spaces. The net effect is that these functions form one equivalence class with the zero function and this is the unique element mapping to zero in the definition of the Fourier transform on $L^1$. Results for the Fourier Cosine and Fourier Sine transform follow as above from the Fourier transform.