3
$\begingroup$

I'm reading through an analysis textbook, and just working through a section on linear functionals. I have the definition: $f$ is a bounded linear functional if $$||f|| = \sup\{|f(x)| : x\in X, ||x|| \le 1\} < \infty,$$ and the proposition that

TFAE:

1. $f$ is bounded;

2. $f$ is continuous;

3. $f$ is continuous at $0$.

Here is my question: it really bothers me, in the definition of a bounded linear functional, that we take just the supremum over all $x$ in the unit ball of the normed space. I know that this is just the definition, but I'm wondering if there is any sort of intuition that would help me to understand why when we find bounds of functions over, say, $R$, we would take the supremum of $|f(x)|$ over all $R$, but with functionals, we take the supremum over just the unit ball?

Another related question that I have is that, in proving $3 \implies 1$ in the proposition above, the book I am reading says:

If $f$ is not bounded, there exists a sequence $x_n \in X$ such that $||x_n|| = 1$ for each $n$, but $|f(x_n)| \to \infty$.

I see that if $||f|| = \infty$, then there must be some sequence $x_n$ such that $|f(x_n)| \to \infty$, $||x_n|| \le 1$ for each $n$. Why can we find such a sequence such that $||x_n|| = 1$?

$\endgroup$
1
$\begingroup$

To answer your second question: Assume that for some $x_n$ in your sequence it holds that $0 < \vert\vert x_n \vert\vert < 1$ (without loss of generality $0 < \vert\vert x_n \vert\vert$, since otherwise $x_n=0$ and $f(x_n)=0$, which cannot be the case for $n$ large enough). Now define a new sequence $y_n = \frac{x_n}{\vert\vert x_n \vert\vert}$ and notice that $\vert \vert y_n \vert \vert = 1$, while $|f(y_n)| = \frac{1}{\vert\vert x_n \vert \vert}|f(x_n)| \geq |f(x_n)| $ by construction of $x_n$ and linearity of $f$. Then of course we will also have $|f(y_n)| \to \infty$.

Now on to your first question: The key point to observe here is that the definition of boundedness for linear functionals is different from the usual definition. In fact, you can check for yourself using an argument similar to the one I gave above (for your second question), that any linear functional (which is not the trivial one mapping everything to 0) which is defined on an unbounded space $X$ is unbounded in the usual sense. This is because as the norm of the function's argument gets bigger, so does the norm of the function's value at that argument.

Therefore, for linear functionals we need a new norm, in place of the usual supremum norm, which in a sense is invariant to the scaling of the argument. Therefore we restrict ourselves only to $x$ with $\vert \vert x \vert \vert = 1$ (indeed I think the fact that in your definition the supremum is taken over $\vert \vert x \vert \vert \leq 1$ might confuse you a bit, but again you can use the usual argument to show that these definitions are equivalent).

Our new norm then is defined as:

$$ \vert \vert f\vert \vert_{OP} = \sup_{x:\vert\vert x\vert\vert=1}|f(x)| = \sup_{x:\vert\vert x\vert\vert \leq 1}|f(x)| $$

Then we call a linear functional bounded, if this norm is $<\infty$ and otherwise unbounded. For linear functionals on finite dimensional spaces this is actually always true, but once you go to infinite dimensional spaces this is no longer the case. But as your results show, this "boundedness" is then exactly equivalent to continuity.

$\endgroup$
3
$\begingroup$

The answer to both of your questions is based on the linearity of $f$. For the first question, notice that, if there is even a single $x$ with $f(x)\neq0$, then by multiplying $x$ by a large positive real number $r$, you get $|f(rx)|=r|f(x)|$, which gets arbitrarily large if you take $r$ large enough. So the only way $f$ could be bounded everywhere (rather than just on the unit ball $\{x:\Vert x\Vert\leq 1\}$) is to be identically $0$.

For the second question, if you have $x$'s with $\Vert x\Vert\leq 1$ and $|f(x)|$ large, then let $y=x/\Vert x\Vert$ (the denominator isn't $0$ because $|f(0)|$ isn't large), and you have $\Vert y\Vert=1$ and $|f(y)|$ is even larger than $|f(x)|$ because $f(y)=f(x)/\Vert x\Vert$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.