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Suppose $g\in L_2([0,1],\lambda)$. I would like to justify that the following map is continuous on $[0,1]$: $$G(u)=\int_{0,u} xg(x)d\lambda(x)$$ The fundamental theorem of calculus requires $xg(x)$ to be continuous, but so far I only know it is in $L_2$. What could be a good justification?

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$G$ is Hölder continuous by Cauchy-Schwarz inequality. –  Davide Giraudo Oct 17 '12 at 13:40

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If $u_1\leq u_2$, we have $$|G(u_1)-G(u_2)|=\left|\int_{(u_1,u_2)}xg(x)d\lambda(x)\right|\leq\int_{(u_1,u_2)} \left|xg(x)\right|d\lambda(x) \\ \leq \int_{(u_1,u_2)} \left|g(x)\right|d\lambda(x)\leq\sqrt{|u_2-u_1|}\cdot\lVert g\rVert_{L^2}.$$

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Thanks! that was straightforward. Isn't it possible to prove the differentiability of G with the same argument? –  david Oct 17 '12 at 14:19
    
We can get, with a better above bounding, that $G$ is Lipschitz, hence is, by Rademacher theorem, almost everywhere differentiable. –  Davide Giraudo Oct 17 '12 at 14:31

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