Norm for continuous linear functionals, newbie questions

Question

Let $E$ be a normed vector space and let $f\colon E \to \mathbb{R}$ be a continuous linear functional. Define the dual norm of $f$ as

$$ \|f\| = \sup_{\|x\|\leq 1} |f(x)|. $$

First question. I have to prove that the supremum defined is the same as we were to find it on the unitary sphere $\|x\|=1$. I fixed a $x$ on the sphere and considered the segment $tx$ with $t \in [0,1]$. So $|f(tx)|=t|f(x)|$, and the values are increasing from $0$ (in which $f$ is $0$) to $1$; repeating for all $x$ and observing that $f$ has max on the sphere, we get the result. Hope it is correct.

Second question. Now I have to prove that

$$ \|f\| = \sup_{\|x\|\leq 1} |f(x)| = \|f\| = \sup_{\|x\|\leq 1} f(x). $$ I don't know how to get rid of modulus, any idea?

Answer 1

Your argument for the first question is essentially right.

If $0 \lt \|x\| \lt 1$ then pick $\lambda = \frac{1}{\|x\|} \gt 1$ and observe that $\|\lambda x\| = 1$. But $|f(\lambda x)| = |\lambda| \,|f(x)| \geq |f(x)|$ with strict inequality if $|f(x)| \neq 0$. Thus, the supremum on the closed unit ball will be attained on the unit sphere.

An important detail:

It is not true that the supremum is a maximum⁽¹⁾ on the unit sphere unless $E$ is finite-dimensional (because then the unit sphere is compact) or, more generally, reflexive⁽²⁾.

As Nate gave you a great hint that should lead you to the goal for the second question, I'll not discuss this part of your question further unless you ask me to.

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⁽¹⁾ For a specific example where the norm of a functional is a supremum but not a maximum, one standard example is to take $E = C[0,1]$ (the space of continuous functions on $[0,1]$ with the supremum norm) and to consider the integral functional $$\varphi: h \mapsto \int_{0}^{1/2} h(t)\,dt - \int_{1/2}^1 h(t)\,dt.$$ It is not difficult to show that $\varphi$ is a functional of norm $1$, and that for every continuous function with supremum $\leq 1$ we have $|\varphi(h)| \lt 1$: Intuitively, this is because a function should be constant equal to $\pm 1$ on $[0,1/2]$ and $[1/2,1]$ (with opposite signs) to maximize $|\varphi(h)|$, however a continuous function will be forced to assume all values in $(-1,1)$ which will lead to the strict inequality $|\varphi(h)| \lt 1$. For details and some remarks on this, you can consult this detailed answer or this thread.

⁽²⁾ It is possible to show (using some functional analytic machinery) that on reflexive spaces the supremum is indeed a maximum. It is a deep theorem due to James that the converse holds: a Banach space is reflexive if and only if every linear functional attains its maximum on the closed unit ball.