# Is this expression positive, negative or both under different regions of positive real numbers?

Please show if this expression is positive, negative or both under different
regions of positive real numbers?

$$g\left(y\right)=\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

Given,

$$x,y,Q\geq0;\;F\left(x,y\right)>0;\;F\left(\infty,y\right)=1;\;\frac{\partial F\left(x,y\right)}{\partial y}\leq0;\;\frac{\partial F\left(x,y\right)}{\partial x}\geq0$$

Please let me know if the question is not well formed or if anything
is not clear.

# Steps Tried,

$$g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2F\left(Q,y\right)\,\left\{ \int_{0}^{Q}dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2QF\left(Q,y\right)\,\int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<0$$
If the below condition holds,
$$\int_{0}^{Q}2\left(Q-x\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<\int_{0}^{Q}2QF\left(Q,y\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$
$$\int_{0}^{Q}2\left(Q-x\right)\,dx>\int_{0}^{Q}2QF\left(Q,y\right)\,dx$$
$$\left|\left(Qx-\frac{x^{2}}{2}\right)\right|_{0}^{Q}>QF\left(Q,y\right)\left|x\right|_{0}^{Q}$$
$$\frac{Q^{2}}{2}>Q^{2}F\left(Q,y\right)$$
$$F\left(Q,y\right)<\frac{1}{2}$$

Alternately,

$$g\left(y\right)\geq\int_{0}^{Q}2Q\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\geq\int_{0}^{Q}\left[2Q-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \right]\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

Am unable to simplify this alternate condition further.

# Related Question:

Inference Regarding Definite Integral of Product of Two Functions

# Steps Tried,

$$g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2F\left(Q,y\right)\,\left\{ \int_{0}^{Q}dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}2\left(Q-x\right)\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2QF\left(Q,y\right)\,\int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\leq\int_{0}^{Q}\left[2\left(Q-x\right)-2QF\left(Q,y\right)\right]\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<0$$
If the below condition holds,
$$\int_{0}^{Q}2\left(Q-x\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx<\int_{0}^{Q}2QF\left(Q,y\right)\,\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$
$$\int_{0}^{Q}2\left(Q-x\right)\,dx>\int_{0}^{Q}2QF\left(Q,y\right)\,dx$$
$$\left|\left(Qx-\frac{x^{2}}{2}\right)\right|_{0}^{Q}>QF\left(Q,y\right)\left|x\right|_{0}^{Q}$$
$$\frac{Q^{2}}{2}>Q^{2}F\left(Q,y\right)$$
$$F\left(Q,y\right)<\frac{1}{2}$$

Alternately,

$$g\left(y\right)\geq\int_{0}^{Q}2Q\frac{\partial F\left(x,y\right)}{\partial y}\,dx-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \int_{0}^{Q}\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

$$g\left(y\right)\geq\int_{0}^{Q}\left[2Q-2\left\{ \int_{0}^{Q}F\left(x,y\right)\,dx\right\} \right]\frac{\partial F\left(x,y\right)}{\partial y}\,dx$$

Am unable to simplify this alternate condition further.