Discussion of the proof that the uniform norm really is a norm

In response to a request on Piazza, I gave a detailed proof that the uniform really is a norm, including some comments and warnings on possible pitfalls along the way. Here (essentially) is what I said.

Let’s have a look at \(\|\cdot\|_\infty\) on the bounded, real-valued functions on \([0,1]\) (but you can also work with bounded functions on any non-empty set you like, and the same proof will work).

We define

\(\ell_\infty([0,1]) = \{f: f \textrm{ is a bounded, real-valued function on }[0,1]\,\}\,,\)

and this is the set/space we will look at. So we set \(Y=\ell_\infty([0,1])\). Then you can check for yourselves that \(Y\) really is a vector space over \(\mathbb{R}\) (using the usual "pointwise" operations for functions). Note that the zero element in this vector space is the constant function \(0\) (from \([0,1]\) to \(\mathbb{R}\)), which I'll denote by \(\mathbf{0}\) here for clarity. (We also used \(Z\) for this function in one of the exercises.)

I'll first give the proof that the uniform norm really is a norm with lots of aside comments (which makes it look long and complicated). Then at the end I'll generate a proof with no comments, so that you can see that it isn't really that long after all.

We now attempt to define the uniform norm, \(\|\cdot\|_\infty\), on the vector space \(Y\).

Let \(f \in Y\).  We wish to define

\(\|f\|_\infty = \sup \{|f(t)|:t \in [0,1]\,\}\,.\)

Notes/comments:

• Before we start to check the three axioms for normed spaces, we first should check that \(\|\cdot\|_\infty\) really makes sense and does define a function from \(Y\) to \(\mathbb{R}^+\). This is sometimes completely obvious, but here we should say something as part of the proof.
• Sometimes in exam questions you are told that you can assume that the "potential norm" does give actually give a function taking values in \(\mathbb{R}^+\). In this case, you can omit this first part of the proof.
• At this point the proofs for \(C[0,1]\) and \(\ell_\infty([0,1])\) differ slightly, because for \(C[0,1]\) we need to appeal to a theorem that tells us that every continuous, real-valued function on \([0,1]\) is bounded. This is part of the Extreme Value Theorem (sometimes called the Boundedness Theorem). Indeed, you can check that \(C[0,1]\) is a linear (vector) subspace of \(\ell_\infty([0,1])\). But for \(\ell_\infty([0,1])\) we don't need to appeal to any theorem at this point.
• Let's get back to the proof! I'll just repeat that first bit!

Let \(f \in Y\).  We wish to define

\(\|f\|_\infty = \sup \{|f(t)|:t \in [0,1]\,\}\,.\)

Since \(f\) is a bounded, real-valued function on the (non-empty) set \([0,1]\), the set

\(\{|f(t)|:t \in [0,1]\,\}\)

is a non-empty, bounded subset of \(\mathbb{R}^+\), and so the supremum of this set is well-defined, and is an element of \(\mathbb{R}^+\). Thus \(\|f\|_\infty\) is well-defined, and \(\|\cdot\|_\infty\) really is a function from \(Y\) to \([0,1]\).

Comments:

• We now check the three axioms for normed spaces, remembering that the first axiom has two separate implications, and we should check these implications separately.
• The rest of this proof works equally well for \(C[0,1]\), though there you have the option of using \(\max\) instead of \(\sup\). We can't use \(\max\) for \(\ell_\infty([0,1])\) because, as we saw in class, without continuity, bounded, real-valued functions on \([0,1]\) might not have a maximum or minimum value there.

First axiom:
Let \(f \in Y\).

Suppose that \(f=\mathbf{0}\), so that \(f(t)=0\) for all \(t \in [0,1]\). Then

\(\|f\|_\infty = \sup \{|f(t)|:t \in [0,1]\,\} = \sup \{0\} = 0\,.\)

Conversely, suppose that \(\|f\|_\infty = 0\). Then \(\sup \{|f(t)|:t \in [0,1]\,\} = 0\), and so for all \(t \in [0,1]\) we have (by definition of \(\sup\), and the fact that modulus is non-negative)

\(0 \leq |f(t)| \leq 0\).

Thus we must have \(|f(t)|=0\) for all \(t \in [0,1]\), and so \(f=\mathbf{0}\).

Comment:

• For this proof, I would actually accept a shorter proof simply stating that a supremum of a non-empty set of non-negative real numbers is \(0\) if and only if all of the real numbers in the set are \(0\) (i.e., the set is actually equal to \(\{0\}\)). The same reasoning is ok for a sum of non-negative real numbers.
• Warning! This shorter argument is not valid for the integral of (for example) a non-negative, Riemann integrable real-valued function, as we discussed in class. So this part of the proof is harder for \(\|\cdot\|\) on \(C[0,1]\): you have to make use of the continuity to help you in that case!
• For the next axiom, I will allow you to assume tacitly, without proof, that if you multiply all of the elements of a non-empty, bounded subset \(A\) of the real numbers by the same non-negative real constant \(c\), then the supremum of the resulting set you get is \(c\) times the old supremum. That is, under these assumptions on \(A\) and \(c\),
  
             \(\sup\{ca:a\in A\} = c \sup A\,.\)
  
  If you have never checked this, you should! (What happens for negative constants \(c\)?)

Second axiom:
Let \(\alpha \in \mathbb{R}\), and let \(f \in Y\). Then we have
 

\(\|\alpha f\|_\infty = \sup\{|\alpha f(t)|:t \in [0,1]\}\)

\(= \sup\{|\alpha| |f(t)|:t \in [0,1]\} = |\alpha| \sup\{|f(t)|:t \in [0,1]\} = |\alpha| \|f\|_\infty\,,\)

 
as required.

Third axiom (triangle inequality):

Comments:

• This is often the hardest rule to check. (On the other hand, this one is relatively easy for \(\|\cdot\|_1\).)
• Actually it isn't that hard here, but there can be problems with the notation, and also with unwarranted assumptions. It can get particularly confusing if you overuse \(t\). I'll comment on this below.

Let \(f,g \in Y\).

Warning! At this point you might attempt the following dubious proof.

[Begin dubious]

\(\|f+g\|_\infty = \sup\{|f(t)+g(t)|: t \in [0,1]\} \leq \sup\{|f(t)|+|g(t)|: t \in [0,1]\}\)

\(=\sup\{|f(t)|: t \in [0,1]\} + \sup\{|g(t)|: t \in [0,1]\} = \|f\|_\infty + \|g\|_\infty.\)

[End dubious]

Actually there is only one false claim in there! What is actually true is that 

\(\sup\{|f(t)|+|g(t)|: t \in [0,1]\} \leq \sup\{|f(t)|: t \in [0,1]\} + \sup\{|g(t)|: t \in [0,1]\}\,,\)

but this is not obvious. Because of this, I also regard the following attempted proof as dubious, even though it has no false statements in it!

[Begin dubious]

\(\|f+g\|_\infty = \sup\{|f(t)+g(t)|: t \in [0,1]\}\)

\(\leq \sup\{|f(t)|: t \in [0,1]\} + \sup\{|g(t)|: t \in [0,1]\} = \|f\|_\infty + \|g\|_\infty.\)

[End dubious]

I recommend the following proof structure instead (but with a warning about overuse of \(t\)).

Let \(t \in [0,1]\). Then, by the definition of \(\|\cdot\|_\infty\), we have \(|f(t)| \leq \|f\|_\infty\) and \(|g(t)| \leq \|g\|_\infty\). Thus we have

\(|f(t)+g(t)| \leq |f(t)| + |g(t)| \leq \|f\|_\infty + \|g\|_\infty\,.\)

This holds for all \(t\) in \([0,1]\), and so we have

\(\|f+g\|_\infty = \sup\{|f(t)+g(t)| : t \in [0,1]\} \leq \|f\|_\infty + \|g\|_\infty\,,\)

as required.  \(\square\)

What about the warning about overuse of \(t\)? Well you might be tempted to write something like

[Begin dubious]
Let \(t \in [0,1]\). Then

\(|f(t)+g(t)| \leq |f(t)| + |g(t)|\)

\(\leq \sup\{|f(t)|: t \in [0,1]\} + \sup\{|g(t)|: t \in [0,1]\} = \|f\|_\infty + \|g\|_\infty\,,\)

and so ...
[End dubious]

But this is unclear because \(t\) is being used for two different things here, which is undesirable. However, the \(t\) in the definition of \(\|\cdot\|\) is a dummy variable, which can be replaced by any other letter.

There is nothing wrong with the following version:

[Begin alternative]
Let \(t \in [0,1]\). Then

\(|f(t)+g(t)| \leq |f(t)| + |g(t)|\)

\(\leq \sup\{|f(s)|: s \in [0,1]\} + \sup\{|g(s)|: s \in [0,1]\} = \|f\|_\infty + \|g\|_\infty\,,\)

and so ...
[End alternative]


Finally, here is an uninterrupted, full proof, as promised!

As mentioned above, I would accept some more concise proofs, as long as all of the main issues were addressed satisfactorily.

Let \(f \in Y\).  We wish to define

\(\|f\|_\infty = \sup \{|f(t)|:t \in [0,1]\,\}\,.\)

Since \(f\) is a bounded, real-valued function on the (non-empty) set \([0,1]\), the set

\(\{|f(t)|:t \in [0,1]\,\}\)

is a non-empty, bounded subset of \(\mathbb{R}^+\), and so the supremum of this set is well-defined, and is an element of \(\mathbb{R}^+\). Thus \(\|f\|_\infty\) is well-defined, and \(\|\cdot\|_\infty\) really is a function from \(Y\) to \([0,1]\).

First axiom:
Let \(f \in Y\).

Suppose that \(f=\mathbf{0}\), so that \(f(t)=0\) for all \(t \in [0,1]\). Then

\(\|f\|_\infty = \sup \{|f(t)|:t \in [0,1]\,\} = \sup \{0\} = 0\,.\)

Conversely, suppose that \(\|f\|_\infty = 0\). Then \(\sup \{|f(t)|:t \in [0,1]\,\} = 0\), and so for all \(t \in [0,1]\) we have (by definition of \(\sup\), and the fact that modulus is non-negative)

\(0 \leq |f(t)| \leq 0\).

Thus we must have \(|f(t)|=0\) for all \(t \in [0,1]\), and so \(f=\mathbf{0}\).

Second axiom:
Let \(\alpha \in \mathbb{R}\), and let \(f \in Y\). Then we have
 

\(\|\alpha f\|_\infty = \sup\{|\alpha f(t)|:t \in [0,1]\}\)

\(= \sup\{|\alpha| |f(t)|:t \in [0,1]\} = |\alpha| \sup\{|f(t)|:t \in [0,1]\} = |\alpha| \|f\|_\infty\,,\)

 
as required.

Third axiom (triangle inequality):
Let \(f,g \in Y\).

Let \(t \in [0,1]\). Then, by the definition of \(\|\cdot\|_\infty\), we have \(|f(t)| \leq \|f\|_\infty\) and \(|g(t)| \leq \|g\|_\infty\). Thus we have

\(|f(t)+g(t)| \leq |f(t)| + |g(t)| \leq \|f\|_\infty + \|g\|_\infty\,.\)

This holds for all \(t\) in \([0,1]\), and so we have

\(\|f+g\|_\infty = \sup\{|f(t)+g(t)| : t \in [0,1]\} \leq \|f\|_\infty + \|g\|_\infty\,,\)

as required.  \(\square\)