$g \in L_1$ such that $g*g \leq g$ nearly everywhere. Show that:$$\int_{\mathbb{R}} g(x) d\lambda_1(x) \leq 1$$
From what is said, I know that:
$$g*g \leq g \implies \int_{\mathbb{R}} (g*g) (x) \ d\lambda_1(x) \leq \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$$
From convolution definition we know that:$$(g*g)(x) = \int_{\mathbb{R}} g(x - y) g(y) \ d(y) \leq g(x)$$
Therefore, our inequality becomes:$$\int_{\mathbb{R}} \left( \int_{\mathbb{R}} g(x - y) g(y) \ d(y) \right) \ d\lambda_1(x) \leq \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$$
I think that since function $g \in L_1$ is Lebesgue integrable, we can change the order of integration and get:
$$\int_{\mathbb{R}} \left( \int_{\mathbb{R}} g(x - y) g(y) \ d\lambda_1(x) \right) \ d(y) \leq \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$$
As it was noticed by Michael Burr in the comment below, $g(y)$ is constant with respect to x, so it is a constant that can be moved outside the integral, so we get:$$\int_{\mathbb{R}} g(y) \left( \int_{\mathbb{R}} g(x - y) \ d\lambda_1(x) \right) \ d(y) \leq \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$$
Then if it was true that $\int_{\mathbb{R}} g(x - y) \ d\lambda_1(x)$ was a constant with respect to $y$, so it could be moved outside the integral also and we would get:$$\int_{\mathbb{R}} g(y) d(y) \cdot \int_{\mathbb{R}} g(x - y) \ d\lambda_1(x) \leq \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$$
My guess is that we want to have that $\int_{\mathbb{R}} g(x - y) \ d\lambda_1(x) = \int_{\mathbb{R}} g(x) \ d\lambda_1(x)$ and then we would obtain the inequality that we were interested in. For that to be true we would need $y$ to be constantly equal to $0$. That would also explain why $\int_{\mathbb{R}} g(x - y) \ d\lambda_1(x)$ would be a constant with respect to $y$... but why would that be the case? Why $y$ would be constantly equal to $0$ in our case?
Any help would be appreciated.