Properties of the image reconstruction filter in rendering
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.