Mathematical relationship between highpass and lowpass filtering Let $g, h_{HP}, h_{LP}: \mathbb{R} \rightarrow \mathbb{R}$ and $G, H_{HP}, H_{LP}$ denote their continuous Fourier transforms under the Fourier operator $\mathcal{F}$. Let $*$ denote the continuous convolution of two functions and $\cdot$ the pointwise multiplication.

Define the shorthand $1: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto 1$ and let $H_{HP} = 1 - H_{LP}$ in the frequency domain, then I would like to show
$$
g * h_{HP} = g - (g * h_{LP}).
$$

In words:


  Filtering an image with a highpass is the same as subtracting the lowpass-filtered image, where the lowpass is the inverted highpass.


There is a proof (from some closed-source course notes) which goes as follows:

\begin{align*}
&amp; g * h_{HP}\\
&amp; = \mathcal{F}^{-1}(G \cdot H_{HP})\\
&amp; = \mathcal{F}^{-1}(G \cdot (1 - H_{LP}))\\
&amp; = g * \mathcal{F}^{-1}(1 - H_{LP}) \quad \text{(The next step is unclear to me)}\\
&amp; = g * (1 - h_{LP})\\
&amp; = g - (g * h_{LP})
\end{align*}

In fact, $\mathcal{F}^{-1}(1 - H_{LP}) = 1 - h_{LP}$ is equivalent to $1 = \mathcal{F}(1)$ by applying $\mathcal{F}$ on both sides, which is a contradiction with the first identity mentioned below.

Questions


  
  Is it true that the proof above lacks details?
  Is my proof below correct/more rigorous?
  


It is known that


$\mathcal{F}(1)=\delta$, from Constant function on MathWorld
$\mathcal{F}^{-1}(1)=\delta$, from Math Stack Exchange: How to prove that inverse Fourier transform of “1” is delta funstion?


I found another approach to possibly prove the proposition above:

First note: If $u, v: \mathbb{R} \rightarrow \mathbb{R}$ are continuous functions and $\int_a^b u(x) dx = \int_a^b v(x) dx$ for every $a, b \in \mathbb{R}$, then $u(x) = v(x)$ for every $x$, i.e. $u=v$.

\begin{align}
&amp;\int_{a}^{b} (g * h_{HP})(x) dx\\
&amp; = \int_{a}^{b} (g * \mathcal{F}^{-1}(1 - H_{LP}))(x) dx \quad \text{(as stated above)}\\
&amp;= \int_{a}^{b} (g * (\delta - h_{LP}))(x) dx \quad \text{(linearity + identity above)}\\
&amp; = \int_{a}^{b} \int_{-\infty}^{\infty} g(\tau) ((\delta - h_{LP})(x - \tau)) d\tau dx\\
&amp; = \int_{a}^{b} \int_{-\infty}^{\infty} g(\tau) (\delta(x - \tau) - h_{LP}(x - \tau)) d\tau dx\\
&amp; = \int_{a}^{b} \int_{-\infty}^{\infty} g(\tau)\delta(x - \tau) - g(\tau)h_{LP}(x - \tau) d\tau dx\\
&amp; = \int_{a}^{b} g(x) - \int_{-\infty}^{\infty} g(\tau)h_{LP}(x - \tau) d\tau dx\\
&amp; = \int_{a}^{b} g(x) - (g * h_{LP})(x) dx\\
&amp; = \int_{a}^{b} (g - (g * h_{LP}))(x) dx
\end{align}

Especially, I am unsure whether the usage of the $\delta$ distribution is rigorous here.
Yes you are right. The following line (excerpted from your post) is wrong, below which I place the corrected version:
\begin{align*}
&amp; g * h_{HP}\\
&amp; = \mathcal{F}^{-1}(G \cdot H_{HP})\\
&amp; = \mathcal{F}^{-1}(G \cdot (1 - H_{LP}))\\
&amp; = g * \mathcal{F}^{-1}(1 - H_{LP}) &amp;\quad \text{(The next step is unclear to me)}\\
&amp; = g * (1 - h_{LP}) &amp;\quad \text{(This line is wrong)}\\
&amp; = g * (\delta(t) - h_{LP}) &amp;\quad \text{(This is the corrected version)}\\
&amp; = g - (g * h_{LP})
\end{align*}

I don't think so if you really need to go deeper than the above corrected proof to show that the inital claim was right. But it's up to you of course.