The Difficulty of Math

I just recently found the following problem on my Google+ page.

2015 - 1

Students or other math enthusiasts like to post problems there. The above seems to be from a text book or a collection of problems. If you cannot read the copy above: You are to prove that every convex function on an open interval is continuous (the second claim is much easier). I vaguely remember that it was on my exercise sheet in the first semester too. Of course, only very few exceptional students can actually solve such problems. Weather or not impossible or at least very difficult problems are useful in teaching is a matter of discussion for a later blog post.

Today, I want to show you why the problem is hard. As stated, it gives the definition of convexity and then immediately asks about the continuity. That looks like a question about limits. In fact, it can be done that way, but only if you are very, very inventive. Once you think geometrical the problem becomes a lot easier.

I won’t go into tow much details here. First of all, a function is convex, if its graph lies underneath any segment joining two points of the graph.

convex1

The reason for this is that there is equality in the definition of convexity if f is a linear function. From that, you immediately see (with just a little bit of effort) that the graph lies in the following shaded area.

convex2

Thus we see that the function is continuous.

Once you have done that, you need to translate your vision back into algebraic relations. Denoting the two lines by g and h, you might be able to find the correct inequalities and write them down so that readers understand them. Some authors even hide their geometrical intuition at this point. But we can also directly translate our geometrical insight.

Take x < t < d. Then by convexity

\(f(t) \le h(t)\)

Moreover, assume f(t) < g(t). Now c < x < t. Thus there is a lambda (easy to write down) with

\(f(x) = f(\lambda c + (1-\lambda)t) \le \lambda f(c) + (1-\lambda) f(t) < \lambda g(c) + (1-\lambda) g(t) = g(x).\)

This is a contradiction to f(x)=g(x). Thus for x < t < d

\(g(t) \le f(t) \le h(t)\)

and continuity follows from a sandwich principle.

To me the play between geometry and algebra is the key to the solution. A direct proof is a lot more difficult.

If you wish you can try the same problem for multi-variate functions f. Then you need to stick a lot more to the algebra underneath the definition of convexity. However, you will indeed follow the geometrical concepts of the one-dimensional case.

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