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When $g(x)$ and $h(x)$ are given functions, can $f(x)^2+(g*f)(x)+h(x)=0$ be solved for $f(x)$ in closed form (at least with some restrictions to $g,h$)? (The $*$ is not a typo, it really means convolution as in $(f*g)(x)=\int\limits_{\mathbb R}f(y)g(x-y)\,dy$)

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even when $g=0$ the equation does not always admits solutions analytic near 0 (take $h=Id$). it seems pretty complicated... –  Glougloubarbaki Nov 29 '11 at 11:40
    
Sounds like an integral equation to me... –  J. M. Nov 29 '11 at 11:45
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If you mean convolution, you should use $(f*g)(x)$ (or $f*g(x)$) instead of $f(x)*g(x)$ which makes (strictly spoken) no other sense then multiplication. –  Dirk Nov 29 '11 at 12:50
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You should probably say what you mean by convolution as well. For example, $\int_{-\infty}^\infty f(x-t)g(t)\,dt$ or $\sum_{k=0}^x f(x-k)g(k)$ or what? –  GEdgar Nov 29 '11 at 22:57
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So the variable $x$ is real? Are the values real? When you say "analytically" do you mean it should be an analytic function? Or is that a vague way of saying "in closed form"? –  GEdgar Nov 30 '11 at 14:54

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