Something has been bothering me for a few days. Here goes. An space is a space of functions consisting of . So essentially, if we can integrate the p-th power of the function and get a finite answer, then the function is in the space. (The careful reader will note that this is dependent not just on p, but on the space and on the measure. Also, since we only care about the integral, functions that differ by a set of measure zero are considered the same function, i.e. equivalence class).

This is a normed space with norm given by . Well, I could go on for awhile about this, but here is what was troubling me. If you define a function , or the p-th power of the p-norm, then this function is continuous on .

I couldn’t seem to do this in any classical straightforward sense. I tried, I really did, with the epsilons and deltas. Luckily, my inability led me to a rather sneaky method (though it assumes knowledge of two other things that I won’t prove). Assume knowledge of the following facts: that if is convex then is convex, every convex function is continuous, and Holder’s Inequality. These facts should be in any standard advanced analysis or measure theory text.

Let and , examine:

So, is convex and hence continuous. I thought that although this relies on some heavier machinery than a direct way, it was much slicker.

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