Daily Definitions 11/14

In this definitions blog, I will explore the supremum norm in greater detail. First, recall the definition of the supremum norm: If f,g:A\rightarrow\mathbb{R}, then we set the supremum norm \mid\mid f\mid\mid_{\infty,A}:\sup\{\mid f(x)\mid:x\in A\}. for the “uniform size” of f and \mid\mid f-g\mid\mid_{\infty,A} is the “uniform distance” between f and g.

Notice that the word “uniform” comes up in this definition. Recall previous concepts of both uniform continuity and uniform convergence. With both of these ideas, the “uniformity” is characterized by a lack of dependence on x. Well, this same idea is also true for the supremum norm.

Extending upon this idea, it is reasonable to entertain the idea of a connection between uniform convergence and the “uniform distance” between functions. Recall the proposition made in class, that f_n\rightarrow f uniformly on A iff \mid\mid f_n-f\mid\mid_{\infty,A}\rightarrow 0.

We proved this statement to be true. However, consider the statement: If f_n\rightarrow f pointwise on A, then \mid\mid f_n-f\mid\mid_{\infty,A}\rightarrow 0.

This statement is not true. Consider the sequence of functions f_n=\frac{x^2+nx}{n}. f_n\rightarrow f point wise, where f(x)=x. However, notice,
$\sup\mid f_n(x)-x\mid=\sup\mid (\frac{x^2}{n}+x)-x\mid=\sup\mid\frac{x^2}{n}\mid$

But \mid\frac{x^2}{n}\mid is unbounded, so this supremum does not exists by the AoC. Thus, we see uniform convergence is required.

Now, it follows that f_n also lack uniform convergence by our first proposition. With this new understanding, we can now use the supremum norm as a nice tool to prove absence of uniform convergence in the future.

Posted on by Connor Kissane
Categories: Definitions

One Response

  1. Jeremy LeCrone says:
    November 26, 2017 at 12:48 pm

    Good Exploration of the definition of uniform norm. Thanks for the post.

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