Applying the norm and metric on a space of continuous functions

Question

I have defined the space of continuous functions (complex trigonometric functions) by the given definitions:

Definition

A normed vector space, $X$, is said to be complete (Banach) iff and only if every sequence is convergent in norm (Cauchy).

Definition

Let $a < b$ be real numbers. Let then $C[a,b]$ denote the set of real-valued continuous functions with domain $[a, b]$.

Definition

Let $f$ be any real-valued function on some domain $\mathscr{D}$, with variable $t$. We can then define the norm of $f$ by $$||f||_\infty = \underset{t\in \mathscr{D}}{\text{sup}} |f(t)|$$

Definition

The metric, defining the distance between two elements $f_1,f_2\in X$, where $X$ is a Banach space (see def. 3, 4), is then measured by $$d(f_1,f_2)=\underset{t \in \mathscr{D}}{max}\big|f_1(t)-f_2(t)\big|$$

However, one thing does not add up.

When I want to apply the metric on two elements, say $f_1= e^{it}-\cos2t$ and $f_2=e^{2it^2}$, it will only work if I integrate the elements over the interval $[a,b]$ and apply the modulus on the result to get a real value.

Does it mean that I have to use the metric

$$ \forall \ \ f_1,f_2 \in C:d_2(f_1,f_2):=\bigg|\bigg(\int_a^b\big(f_1(t)-f_2(t)\big)^2 dt\bigg)^{\frac{1}{2}}\bigg|$$

and therefore go over the $L^2$ space instead? If yes, then it cannot be Hilbert space, because the elements $f_k$ are nonlinear continuous trigonometric functions on $\mathbb{C}^2$, and are not orthogonal to one another. So this would be $L^2$ "without" an inner product. What would the right name of the vector space be in this case?

Thanks

Answer 1

You have misquoted the definitions.

Let's start with

Definition

A normed vector space, $X$, is said to be complete (Banach) iff and only if every sequence is convergent in norm (Cauchy).

No. A Banach space is a complete normed vector space. But "complete" means that every Cauchy sequence converges. Where a Cauchy sequence is a sequence $\{x_n\}$ that satisfies the Cauchy condition $$\forall \epsilon > 0, \exists N, \forall n,m > N, \|x_n - x_m\| < \epsilon$$

"Cauchy" has to modify "sequence" here (not anything else) to make any sense. You cannot just throw it in at the end as an afterthought.

There is also a problem with this definition:

Definition

Let $f$ be any real-valued function on some domain $\mathscr{D}$, with variable $t$. We can then define the norm of $f$ by $$\|f\|_\infty = \underset{t\in \mathscr{D}}{\text{sup}} |f(t)|$$

That only defines a norm when $f$ is a bounded function. For instance $\mathscr D = (0,\infty)$ is a domain, and $f(x) = \frac 1x$ is a function on it. But $ \underset{t\in \mathscr{D}}{\text{sup}} |f(t)| = \infty$, which is not an allowed value for a norm.

While "bounded" is a necessary condition on $f$, "real-valued" is not. Notice in the definition you are not taking the supremum over values $f(t)$, but rather over values $|f(t)|$. When $f$ is complex-valued, $|f(t)|$ is still only real-valued, so the definition still works.

Now all functions in $C[a,b]$ are bounded, so this shouldn't affect the rest of your post

Definition

The metric, defining the distance between two elements $f_1,f_2\in X$, where $X$ is a Banach space (see def. 3, 4), is then measured by $$d(f_1,f_2)=\underset{t \in\mathscr{D}}{max}\big|f_1(t)-f_2(t)\big|$$

This is okay, but I just want to point out that "Banach" is unnecessary. This is how distance is defined in any space with the supremum (aka uniform) norm. More generally, in any space with a norm $\|\cdot\|, d(f_1, f_2) = \|f_1 - f_2\|$.

When I want to apply the metric on two elements, say $f_1= e^{it}-\cos2t$ and $f_2=e^{2it^2}$, it will only work if I integrate the elements over the interval $[a,b]$ and apply the modulus on the result to get a real value.

This is false. $\|f\|_\infty$ is a valid norm and works just fine for those two functions. You haven't specified what interval $[a,b]$ you are using in this example, but I'll use the obvious $[0,2\pi]$. But it is still valid no matter what interval is used.

$$\begin{align}d(f_1,f_2) &= \|f_1 - f_2\|_\infty\\&=\sup_{0 \le t \le 2\pi} |f_1(t) - f_2(t)|\\&=\sup_{0 \le t \le 2\pi} |e^{it} - \cos 2t - e^{2it^2}|\\&=\sup \{ |e^{it} - \cos 2t - e^{2it^2}| : 0 \le t\le 2\pi\}\end{align}$$

Note that $|e^{it} - \cos 2t - e^{2it^2}|\le |e^{it}| + |\cos 2t| + |e^{2it^2}| \le 1 + 1 + 1 = 3$ for all $t$.

Thus every element of the set $\{ |e^{it} - \cos 2t - e^{2it^2}| : 0 \le t\le 2\pi\}$ is $\le 3$ in value. Any set of real numbers that is bounded above has a supremum. The value of that supremum may not always be easy to calculate, but it still exists.