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In cases (a) and (c) [where it was proven that such a number exists for a continous $f$ on $\textbf{R}$], let $g(x)$ be the minimum distance from $(x,0)$ to a point on the graph $f$. Prove that $g(y)\leq g(x)+|x-y|$, and conclude that $g$ is continous.

The answer from the answer book

By definition, $g(x) = \sqrt{(f(z))^2+(z-x)^2}$ for some $z$ in $[a,b]$. Now $\sqrt{(f(z))^2+(z-y)^2} \leq \sqrt{(f(z))^2+(z-x)^2}+|z-y|$ for all $z$. So $g(y)$, the minimum of all $\sqrt{(f(z))^2+(z-y)^2}$, is less than or equal to $|z-y|+$ the minimum of all $\sqrt{(f(z))^2+(z-x)^2}$, which is $g(x)+|x-y|$. Since $|g(y)-g(x)|<|y-x|$ it follows that $g$ is continous (given $\varepsilon>0$, let $\delta = \varepsilon$).

I understand why $\sqrt{(f(z))^2+(z-y)^2} \leq \sqrt{(f(z))^2+(z-x)^2}+|z-y|$ for all $z$ and I see how the continuity of $g$ would follow from the conclusion. But I don't get why

the minimum of all $\sqrt{(f(z))^2+(z-y)^2}$, is less than or equal to $|z-y|+$ the minimum of all $\sqrt{(f(z))^2+(z-x)^2}$

and why this is supposed to be $g(x)+|x-y|$.

Can someone help me here?

Edit: I've now figured out the following things: if for some $z_0$ we have a minimum of $\sqrt{(f(z))^2+(z-y)^2}$, this means that $$ g(y) \leq \sqrt{(f(z))^2+(z-y)^2} \leq \sqrt{(f(z))^2+(z-x)^2} +|y-z| $$ for all $z$. So we can simply pick a $z=u$ such that $\sqrt{(f(u))^2+(u-x)^2} = g(x)$, so then we have $$ g(y) \leq g(x)+|u-y| $$ Does anyone now know how to turn that $|u-y|$ into $|x-y|$?

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So in the end I solved this using a pretty basic method: the triangle inequality (proven in problem 4-9 b) in the book).

So the whole situation can be described with a picture:

distances

We see that the points $(x,0)$, $(y,0)$ and $(z, f(z))$ represent a triangle. And according to the triangle inequality we have, for every $z$ $$ \sqrt{(f(z))^2+(y-z)^2} \leq \sqrt{(f(z))^2+(x-z)^2}+|x-y|. $$

Now for some $z_0$ we have $g(y) = \sqrt{(f(z_0))^2+(y-z_0)^2} \leq \sqrt{(f(z))^2+(y-z)^2}$ for all $z$. This means that for all $z$ $$ g(y) \leq \sqrt{(f(z))^2+(x-z)^2} + |x-y|. $$ Now we simply choose a number $u$ such that $$ g(x) = \sqrt{(f(u))^2+(x-u)^2} $$ So we have $g(y) \leq g(x) + |x-y|$.

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