How to Prove Inequalities for Infima and Suprema

The title does not state which inequality needs proving. A standard result that fits the topic is this: if $f$ and $g$ are bounded real-valued functions on a nonempty set $X$, then the difference between their infima is controlled by the supremum of their pointwise absolute differences. The same argument also gives useful inequalities for sums and suprema.

The basic inequality

For bounded functions $f,g:X\to\mathbb{R}$, define

$$\delta=\sup_{x\in X}|f(x)-g(x)|.$$

Then

$$|\inf_{x\in X}f(x)-\inf_{x\in X}g(x)|\leq \delta.$$

In words, if $f$ and $g$ never differ by more than $\delta$, their infima cannot differ by more than $\delta$. The corresponding statement for suprema is

$$|\sup_{x\in X}f(x)-\sup_{x\in X}g(x)|\leq \delta.$$

The assumptions ensure that the infima and suprema are finite real numbers. If you work with extended real values, additional care is needed when subtracting infinities.

Proof for infima

By definition of $\delta$, for every $x\in X$,

$$|f(x)-g(x)|\leq\delta.$$

Thus $f(x)\geq g(x)-\delta$ for every $x$. Since $g(x)\geq\inf_X g$ for every $x$, it follows that

$$f(x)\geq\inf_X g-\delta\qquad\text{for every }x\in X.$$

The number $\inf_X g-\delta$ is therefore a lower bound for the values of $f$. The greatest lower bound of those values must be at least as large, so

$$\inf_X f\geq\inf_X g-\delta.$$

Interchanging $f$ and $g$ gives $\inf_X g\geq\inf_X f-\delta$. Together, these inequalities say that each infimum is no more than $\delta$ below the other, which is exactly

$$|\inf_X f-\inf_X g|\leq\delta.$$

For the supremum version, use $f(x)\leq g(x)+\delta$ and then reverse the roles of the functions. This yields $\sup_X f\leq\sup_X g+\delta$ and $\sup_X g\leq\sup_X f+\delta$.

Worked example

Take $X=[0,1]$, $f(x)=x$, and $g(x)=x+0.4$. For every $x$,

$$|f(x)-g(x)|=0.4,$$

so $\delta=0.4$. The infima are $\inf_X f=0$ and $\inf_X g=0.4$. Hence

$$|\inf_X f-\inf_X g|=|0-0.4|=0.4=\delta.$$

The bound is attained in this example. The suprema are $1$ and $1.4$, and their absolute difference is also $0.4$.

Related inequalities for sums

The same pointwise reasoning gives standard bounds for two bounded functions on the same nonempty set:

$$\inf_X f+\inf_X g\leq\inf_X(f+g),$$

and

$$\sup_X(f+g)\leq\sup_X f+\sup_X g.$$

For the first inequality, every $x$ satisfies $f(x)\geq\inf_X f$ and $g(x)\geq\inf_X g$. Adding gives $f(x)+g(x)\geq\inf_X f+\inf_X g$; take the infimum over $x$. For the second, add the pointwise upper bounds and take the supremum.

These need not be equalities. For example, on $X=\{1,2\}$, let $f(1)=0$, $f(2)=1$, $g(1)=1$, and $g(2)=0$. Then $\inf_X f+\inf_X g=0$, while $f(x)+g(x)=1$ at both points, so $\inf_X(f+g)=1$. The two functions attain their minima at different points.

Common mistakes

• Treating an infimum as a minimum. An infimum need not be attained. The proof uses lower-bound properties, not a point where the function reaches its infimum.
• Reversing an inequality when taking an infimum. If $a(x)\leq b(x)$ for every $x$, then $\inf_X a\leq\inf_X b$. For suprema, the same pointwise order gives $\sup_X a\leq\sup_X b$.
• Assuming the sum inequality is always an equality. Equality can fail when the functions reach their extrema at different inputs.
• Forgetting the domain. Infima and suprema here are taken over the same nonempty set $X$. If the domains differ, first specify how the functions are being compared.

In a metric-space setting, the same result says that the map taking a bounded function to its infimum changes by at most the uniform distance between functions. In particular, if the uniform distance is defined by $d_\infty(f,g)=\sup_{x\in X}|f(x)-g(x)|$, then $|\inf_X f-\inf_X g|\leq d_\infty(f,g)$. No monotonicity assumption is needed. If a different inequality was intended, its exact expression is necessary to prove that particular statement.