Meaning of frequency in DFT

Question

I am trying to get a better understanding of DFT of an image. Assume that we have an image of size $N\times N$, denoted as $f(m, n), m, n = 0, 1, \ldots, N$, and its Fourier transform $F(k, l), k, l = 0,1, \ldots, N-1$. It is often said that $F(N-1, N-1)$ represents the highest frequency. But what exactly does it mean?

If I was transforming a $1D$ signal, the "base" functions (I quote "base" because from my understanding of linear algebra DFT is not really a change of basis) are $e^{-i2\pi \frac{km}{N}}$ where $m$ is index of signal in space domain and $k$ is index in Fourier domain. For $k=N-1$ I get $F(N-1) = f(0)e^{0}+f(1)e^{-i2\pi \frac{N-1}{N}} + f(2)e^{-i4 \pi \frac{N-1}{N}} + \ldots f(N-1)e^{-i(N-1) \pi \frac{N-1}{N}}$. Now, some of the "base" functions in $F(N-1)$ are contained in other $F(k), k < N-1$, but definitely not the last one $e^{-i(N-1) \pi \frac{N-1}{N}}$. So does the "highest frequency" mean that the range of the arguments of the exponentials in $F(N-1)$ is larger than in any other $F(k), k<N-1$?

Answer 1

It might be helpful to think of the DFT as correlating the signal $f(n)$ with a set of “basis” functions, these being complex exponentials of the form: $$W_k(n) = e^{-i2\pi \frac{k}{N}n}, \quad 0\leq n<N$$ where $\frac{k}{N}$ is the discrete frequency used to build a particular function of this set.
For example, the 3rd bin of the DFT of $f(n)$ contains the result of correlating $f(n)$ with a complex exponential of frequency $\frac{3}{N}$.

The higher $k$ is, the higher the base exponential’s frequency is. When the DFT gets to the case where $k = N-1$, it simply means it’s correlating $f(n)$ with the complex exponential of highest frequency allowed by the DFT size $N$.