The Kramer-Kronig (KK) relations state that for ALL linear, causal and stable systems, the real and imaginary components of the Fourier Transform are related by the Hilbert Transform. For $h(t)$ and $H(f)$ as a Fourier Transform pair:
$$h(t) \leftrightarrow H(f)$$
If $h(t)$ is one-sided (causal) and stable (which means all poles of the Laplace Transform are in the left half plane), then $H(f)$ MUST be complex as:
$$H(f)= H_R(f)+jH_I(f)$$
Where $H_R$ and $H_I$ refer to the real and imaginary components of $H(f)$ respectively.
And further (as given by KK) the imaginary component $H_I(f)$ is related to the real component $H_R(f)$ via the Hilbert Transform as follows:
$$H_I(f) = -\mathscr{H}\{H_R(f)\} \tag{1} \label{1}$$
The above relationship is easy to show quite intuitively by decomposing $h(t)$ into even and odd waveforms and following the Fourier Transform properties for such waveforms (the transform of an even real waveform in time will be real in frequency while the transform of an odd real waveform in time will be imaginary in frequency). For further explanation please see this write-up. From that we deduce that equation \ref{1} always holds as long as the system is linear, causal and stable (but not otherwise).
Meanwhile the Bode relations state that for ONLY minimum phase linear, causal and stable systems the magnitude and phase are also related by the Hilbert Transform as follows:
$$\phi(f) = -\mathscr{H}\{|H(f)|)\} \tag{2} \label{2}$$
Interesting! And why the restriction to being minimum phase since we already know the real and imaginary components are related by equation \ref{1}??
This too is straight-forward and can be shown to be intuitive assuming we have a knowledge of poles and zeros and how that relates to minimum phase systems (specifically that all poles and zeros must ber in the left half plane to be minimum phase, and all poles must be in the left half plane to be causal and stable; see DSP.SE#2241 for more details), and as long as we believe what has been told to us by KK. First we can see how the relationship in equation \ref{2} is established, and then from that ask if we still meet the conditions established for KK.
Equation \ref{2} comes from describing the complex $H(f)$ in polar form (magnitude and phase) as follows:
$$H(f) = |H(f)|e^{j\phi(f)}$$
If we were to take the natural log of $H(f)$, this would allow us to isolate the magnitude and phase components, but note that it is also describing another system, which I will refer to as $G(f)$:
$$G(f) = \ln(H(f)) = \ln(|H(f)|e^{j\phi(f)}) = \ln(|H(f)| + j\phi(f)$$
Interesting again! We have created a system with real and imaginary components:
$$G(f) = G_R(f) + j G_I(f)$$
Where $G_R(f) = \ln(|H(f)|$ and $G_I(f) = \phi(f)$
So IF $G_R(f)$ is stable and causal, then we have from KK the relationship similar to equation \ref{1}:
$$G_I(f) = -\mathscr{H}\{G_R(f)\} $$
Resulting in equation \ref{2}!
So here's the catch: We took the natural log of $H(f)$, and $H(f)$ specifically is the evaluation of the Laplace Transform $H(s)$ with $s=j\omega$. The system is characterized by $H(s)$ and notably by the location of its poles and zeros on the complete s-plane. If the poles are in the left-half plane and the system is causal, then it is also stable. If the zeros are also in the left-half plane, the system is minimum phase. If we consider the other system $G(s)$ given by $G(s)=\ln(H(s)$); the locations where $H(s)$ has poles will also be poles in $G(s)$ (as $\lim_{x\rightarrow\infty}\ln(x) = \infty$), but for the locations where $H(s)$ has zeros, those will become poles in $G(s)$! (since $\lim_{x\rightarrow 0}\ln(x) = \infty$). So there you have it! If $H(s)$ did have any zeros in the right half plane (meaning it is NOT minimum phase), then $\ln(H(s)$ will have poles in the right half plane and therefore is not a causal stable system, so doesn't meet the conditions for KK.
Note: I used Laplace and s-plane describing continuous-time systems, but the explanation above holds for discrete time systems as well using $H(z)$ and $G(z)$ and inside/output the unit circle on the z-plane instead of in the left half / right half plane of the s-plane.
