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6
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_1,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2.
Edit: This argument doesn't work without additional assumptions; see the comments.
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5
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_1,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2.
Edit: This argument doesn't work; see the comments.
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4
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_0,y_1)$ F(x_1,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2.
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3
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_0,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq 2M(F(x_0,y_0)-F(x_1,y_1))$M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$.2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2.
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2
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_0,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$2M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$.
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1
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You can use the triangle inequality to solve this by looking at each coordinate separately.
$F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_0,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$.
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