# Extended real-valued function

1. Jun 28, 2010

### wayneckm

Hello all,

Recently I came across the following statement:

What happens when a convex function f achieves the value −1 at some point xo? Usually, a degenerate behaviour occurs. For instance, suppose that f is defined on R, and f(0) =
− infinity. If f(1) is finite (say), then one must have f(x) = −1 for all 0 <= x < 1 and f(x) = +infinity for all x > 1.

Apparently there is no restriction on the function characteristics, e.g. continuity, on f, why is it a "MUST"? If f is continuous (is this allowed for extended real-valued function?), it seems this is not a "MUST".

Or please kindly advise me on the definition of extended real-valued function as well as its characteristics.

Thanks very much.

2. Jun 28, 2010

### Hurkyl

Staff Emeritus
Surely there's at least one typo in there?

Anyways, I don't think the term "convex function" really makes sense in this context.

3. Jun 28, 2010

### wayneckm

Oops, I think I omiited the assumption of convexity in f.

For $$0 \leq \alpha < 1$$, by convexity, we have $$f(\alpha) \leq \alpha f(0) + (1-\alpha) f(1) \; \Rightarrow \; f(\alpha) \leq -\infty$$, so we deduce that $$f(x) = -\infty \quad \forall x \in [0,1)$$

Similarly we can prove $$f(x) = +\infty \quad \forall x \in (1,+\infty)$$ otherwise it would violate the assumption of convexity, in particular at $$f(1)$$ which is finite.

4. Jun 28, 2010

### Hurkyl

Staff Emeritus
Alas, convexity is violated anyways: we have now shown the right hand side of
$$f(1) \leq (1/2) f(0) + (1/2) f(2)$$​
must be undefined. (being of the form $(+\infty) + (-\infty)$)

5. Jun 28, 2010

### wayneckm

So it seems the results rely on whether one define the operation $$+\infty + -\infty$$? And are we free to define this kind of operation in the extended real number system?

6. Jun 28, 2010

### Hurkyl

Staff Emeritus
In the extended real number system, $(+\infty) + (-\infty)$ is undefined.

There is, of course, nothing stopping you from defining a different number system in whatever way you like whose numbers are extended real numbers.

More fruitful is to come up with an appropriate definition for the extended reals, rather than try to force a definition meant for standard reals to work.

Now that I think more about it, I suspect "convex function" really is a reasonable notion for extended real numbers. I would guess its definition would be equivalent to:
f is convex iff the set of all real numbers (x,y) satisfying $y \geq f(x)$ is a convex subset of the plane​
(One could write an equivalent definition in terms of the algebraic identity you used, but with ad hoc additions to treat the cases where the function is somewhere infinite)