“How to measure the distance between two functions?”

Just as you can have a sequence

• of number in $\R$
• of vectors in $\R^n$
• of points in a general metric space $X$.

You can have a sequence of functions. $f_n(x)$

Let $V = \{f: f: [0,1] \to \R\}$ be the space of bounded functions from $[0,1]$ to $\R$, we will consider various ways to measure size of a function

• sup norm: $ \| f \|_\infty = \sup \{ f(x) : x \in [0,1]\} $
• $L^1$ norm $ \| f\|_1 = \int |f(x)| dx $
• $L^2$ norm $ \| f\|_2 = (\int |f(x)|^2 dx)^{1/2}$
• in general, for $1 \leq p < \infty$, we have $L^p$ norm.

The different norms equip the vector space $V$ with different topologies. This is different from the finite dimensional vector space case.

In sport, sometimes you measure the score using the best score (like freestyle skiing in the winter Olympics); sometimes you use the average score of several try (like Tour de France, you add up the points in each leg); this is like sup or $L^1$ norm.

In this class, we will consider sup norm, and we define distance as $$ d_\infty(f,g) = \| f-g\|_\infty $$

We say a sequence of functions $f_n$ converge to $f$ uniformly, if $\lim_n d_\infty(f_\infty, f) = 0$.

Q: can you find a metric on the space of functions $V$ so that metric convergence means pointwise convergence? (No, you but you can still define a topology on $V$ so that convergence in that topology means pointwise convergence)

In general, if $M = \{f| f: X \to Y, f(X) \text{bounded}\}$ is the space of maps with bounded images, then for any $f, g \in M$, we can define $d_\infty(f,g) = \sup_{x} d_\infty(f(x), g(x))$.

• The running bump $f_n = 1_{[n,n+1]}(x)$, which is 1 on $[n,n+1]$ and $0$ elsewhere. They converge to $0$ pointwise, but not uniformly.
• The shrinking and rising bump $f_n = n 1_{(0,1/n)}$ converge to $0$ pointwise, but not uniformly.

Thm: If $f_n: \R \to \R$ are continuous and bounded, and $f_n \to f$ uniformly, then $f$ is continuous.

Proof: we need to show that, for any $x \in \R$, for any $\epsilon>0$, there is $\delta > 0$, such that for any $x'$ with $|x' - x| < \delta$, we have $|f(x') - f(x)|<\epsilon$.

First, we choose $n$ large enough, such that $d_\infty(f_n, f) < \epsilon/3$. Then, for $f_n$ and $x$ and $\epsilon/3$, we find $\delta$, such that any $|x'-x|<\delta$ implies $|f_n(x) - f_n(x')|<\epsilon/3$. Finally, if any $|x'-x|<\delta$, then $$ |f(x) - f(x')| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(x')| + |f_n(x') - f_n(x)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon. $$ Done

• Devil's staircase
• power series
• Weierstrass M-test