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I'm stuck with this problem:

$$ \lvert\max_a f(a) - \max_a g(a)\rvert \leq \max_a \lvert f(a) - g(a)\rvert $$ for any $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$.

I would like to understand if the inequality is true, and, if this is the case, how to show it.

I tried with different instances of $f$ and $g$, and seems to be true. I tried to show by contradiction, but I'm not able to.

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  • $\begingroup$ Basically, triangle inequality should prove intuitive. $\endgroup$ Jan 3 '17 at 12:29
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First note that a function $f:\mathbb{R}\to\mathbb{R}$ does not necessarily have a maximum (even if it continuous). It can be unbounded, e.g. $f(x) =x$, or bounded above without attaining the maximum, e.g. $f(x) = x^2/(1+x^2)$.

If $\max_a f(a)$, $\max_a g(a)$ and $\max_a \lvert f(a) - g(a)\rvert $ all exist then you can argue as follows: The maximum of $f$ is attained at some point $b$. Then $$ \max_a f(a) - \max_a g(a) = f(b) - \max_a g(a) \le f(b) - g(b) \leq \lvert f(b) - g(b)\rvert \leq \max_a \lvert f(a) - g(a)\rvert $$ and we have $$ \max_a f(a) - \max_a g(a) \leq \max_a \lvert f(a) - g(a)\rvert \, . $$ The same inequality holds with $f$ and $g$ interchanged, and therefore $$ \lvert\max_a f(a) - \max_a g(a)\rvert \leq \max_a \lvert f(a) - g(a)\rvert \, . $$


If $f$ and $g$ are only assumed to be bounded then a similar argument shows that $$ \lvert\sup_a f(a) - \sup_a g(a)\rvert \leq \sup_a \lvert f(a) - g(a)\rvert \, . $$

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Suppose without loss of generality that $\max\limits_a f(a)>\max\limits_a g(a)$. Then we have:

$\max\limits_a f(a)-\max\limits_a g(a)|=\max\limits_a f(a)-\max\limits_a g(a)\leq \max\limits_a(f(a)-g(a))\\\leq \max\limits_a|f(a)-g(a)|$

Indeed $\max\limits_a f(a)-\max\limits_a g(a)\leq \max\limits_a(f(a)-g(a))$, because:

$\max\limits_a(f(a)-g(a))+\max\limits_ag(a)\geq f(x)-g(x)+g(x)=f(x)\ \forall x \Rightarrow \\ \max\limits_a(f(a)-g(a))+\max\limits_ag(a)\geq \max\limits_a f(a)$

by taking supremum, the maximum of this case, for all $x$ in both sides of the inequality.

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We may assume $\sigma:=\sup_x|f(x)-g(x)|<\infty$. Then for all $x$ the following holds: $$f(x)=g(x)+\bigl(f(x)-g(x)\bigr)\leq g(x)+\bigl|f(x)-g(x)\bigr|\leq g(x)+\sigma\leq\sup_y g(y)+\sigma\ .$$ This allows to conclude that $\sup_x f(x)\leq \sup_y g(y)+\sigma$, and due to symmetry we also have $\sup_y g(y)\leq \sup_x f(x)+\sigma$. Altogether the claim follows.

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