Properties of the image reconstruction filter in rendering Let $f$ denote the measurement contribution function for paths of length $k\in\mathbb N$, i.e. $$f(x)=g(x_0\leftrightarrow x_1)W_{\text e}(x_1\to x_0)t_k(x_0,\ldots,x_k)L_{\text e}(x_k\to x_{k-1}),$$ where $$t_k(x)=\prod_{i=2}^kg(x_{i-1}\leftrightarrow x_i)f_{\text s}(x_i\to x_{i-1}\to x_{i-2}),$$ and $h_j$ denote the image reconstruction filter of the $j$th pixel so that the measurement $I_j$ of the $j$th pixel color is given by $$I_j=\int h_jf\:{\rm d}\mu.$$

What can we assume know about $h_j$ in general? One thing is that it will only depend on first two vertices of a path.

There are two properties which I really would like to being able to assume: (a) $c:=\sup_j\left\|h_j\right\|&lt;\infty$ (b) $\int|g_j|^2\:{\rm d}\mu\le\alpha\int|f_0|^2\:{\rm d}\mu$ for some $\alpha\ge0$, where $g_j:=1_{\{\:h_j\:&gt;\:0\:\}}\left(h_jf-\frac1{\mu\left(\left\{h_j&gt;0\right\}\right)}I_j\right)$ and $f_0:=1_{\{\:f\:\ne\:0\:\}}\left(f-\frac1{\mu\left(\left\{f\ne0\right\}\right)}\int f\:{\rm d}\mu\right)$.

However, I know almost nothing about the mathematical properties of $h_j$ and so it's hard for me to verify if these properties hold. I guess that generally $h_j\ge0$, but I'm even unsure about that. Maybe even $h_j\in[0,1]$? It would be great if someone could answer if (a) and/or (b) hold or link me to a reference.
I am replying to the last edit of the question.

You can show that $\|h_j * f \|^2_2 \leq \|h_j\|^2_1\|f\|^2_2$ from Young's theorem. Provided that  $\|h_j\|_1 \leq 1$ you have: $\|h_j * f \|^2_2 \leq \|f\|^2_2$. Additionally for $g_j$ I believe you want to clamp the indicator function to $\{h_j &gt; 0 \wedge f &gt; 0\}$. Then $\int_{\{h_j&gt;0 \wedge f&gt;0\}} a\leq \int_{\{f&gt;0\}}a$ (since the integrands are always non-negative: $a\geq 0$). This basically proves that the inequality holds for the left terms of the quadratic expansions, the middle terms are both negative, so those may be ignored. The only thing left to prove is: $(\frac{1}{\mu(\{h_j&gt;0 \wedge f&gt;0\})}\int h_jf)^2 \leq (\frac{1}{\mu(\{f&gt;0\})}\int f)^2$, which is the case if $h_j \leq 1$ everywhere. Basically, as long as your filter $h_j$ preserves the minimum and maximum of the image, you're good (this is not an unreasonable requirement for a filter, note however, that there are filers that do not preserve min-max).

Edit: My bad, the middle terms may not be ignored (at least the one on the right of the inequality). Which means you are left to prove that:

$$\frac{(\int h_jf)^2}{\mu(\{h_j&gt;0 \wedge f&gt;0\})} \geq \frac{(\int f)^2}{\mu(\{f&gt;0\})}$$, which will be a problem.

I guess one has to use tighter bounds to relate the mixed term with the squared terms.

Edit 2:

Let's consider the expression as a whole then:

$$\int \left(h_jf - \frac{\int h_jf}{\mu(\{h_j &gt; 0 \wedge f &gt; 0\})} \right)^2 \leq \alpha \int \left(f - \frac{\int f}{\mu(\{f &gt; 0\})} \right)^2$$
$$ 0 \leq \alpha\int (f)^2 - \int (h_jf)^2 +\alpha\frac{(\int f)^2}{\mu^2(\{f &gt; 0\})} - \frac{(\int h_jf)^2}{\mu^2(\{h_j &gt; 0 \wedge f &gt; 0\})} + \frac{2(\int h_jf)^2}{\mu(\{h_j &gt; 0 \wedge f &gt; 0\})} - \alpha\frac{2(\int f)^2}{\mu(\{f &gt; 0\})}$$
$$ 0 \leq \alpha\left(\int (f)^2 - \frac{(1-2\mu(\{f &gt; 0\}))(\int f)^2}{\mu^2(\{f &gt; 0\})}\right) - \left( \int (h_jf)^2 - \frac{(1-2\mu(\{h_j &gt; 0 \wedge f &gt; 0\}))(\int h_jf)^2}{\mu^2(\{h_j &gt; 0 \wedge f &gt; 0\})}\right) $$

$$0 \leq \left(\int(\sqrt{\alpha}f)^2-\int (h_jf)^2\right) - \left(\frac{\alpha(1-2\mu(\{f &gt; 0\}))(\int f)^2}{\mu^2(\{f &gt; 0\})}  - \frac{(1-2\mu(\{h_j &gt; 0 \wedge f &gt; 0\}))(\int h_jf)^2}{\mu^2(\{h_j &gt; 0 \wedge f &gt; 0\})}\right) $$

$$0 \leq \left(\int(\sqrt{\alpha}f)^2-\int (h_jf)^2\right) - \left(a \left(\int \sqrt{\alpha}f\right)^2 - b\left(\int h_jf\right)^2\right)$$

Note also that $\left(\int u\right)^2 \leq \int u^2$. Then with an appropriate $\alpha$ and $h_j$ bounded it should hold. You may want to study the $a,b$ tems also.