Supremum & Infimum with Functions

Question

I found this:

$\sup (f(x)+g(x)) \leq \sup f(x)+\sup g(x)$

But what about $\sup (f(x)+g(x)) = \sup f(x)+\sup g(x)$ ?

With sets we know that

$\sup (A+B) = \sup A+\sup B$

Why this does not hold with functions?

Answer 1

With sets we know that $$\sup (A\cup B) = \sup A+\sup B$$

No we don't. Take $A=B=[0,1]$

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Also, for functions, take

$$f(x)=\begin{cases}1&x<0\\0&x\geq 0\end{cases}$$

and

$$g(x)=\begin{cases}0&x<0\\1&x\geq 0\end{cases}$$

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Or, in fact, take any function you want, $f(x)$. Now, define $g(x)=1-f(x)$.

Then, $\sup(f(x)+g(x)) =\sup 1 = 1$, while $\sup f(x)$ can be just about anything (and so can $\sup g(x)$).

Answer 2

I assume that you invoke the property $sup(A+B) = supA + supB$, which is true. Well, intuitively, think in terms of the trivial case, when the supremum is the maximum. $max(A+B) = maxA + maxB$, this should be clear, since when additing the elemnts of $A$ to the elements of $B$ you also add their maxima. In the case of the functions $f$ and $g$, think what happens if $f(x)=x$ and $g(x)=1-x$; you will find that the equality does not hold.

An other clearer example is taking $f(x)=1+sinx$ and $g(x)=1-sinx$, then $(f+g)(x) = 1$, but $supf + supg = 1 + 1 = 2$!