If $\phi: X \rightarrow \mathbb{R} $ is a bounded linear functional, then the image $\phi(X)$ of $\phi$ is either $\{ 0\}$ or $\mathbb{R}$

Question

I am not entirely sure how to go about this question.

We have that, if the argument of the linear functional is zero, the linear functional is zero (by the definition), but I am not too sure how to go about the proving wheter the image of a nonzero argument is the real line or not.

Prove or disprove: If $\phi: X \rightarrow \mathbb{R} $ is a bounded linear functional, then the image $\phi(X)$ of $\phi$ is either $\{ 0\}$ or $\mathbb{R}$.

My way of thinking was let $X = l_1$ and then let us use the sequence $x= (1, \frac{1}{2}, \frac{1}{4}, ..., \frac{1}{2^n})$. Then, because our linear functional is bounded, it must be the case that $|f(x)|\leq 2C$, where C is a constant. However, then we have the image is not the real line but the interval between $[-2C,2C]$. Is this a decent proof or am I barking up the wrong tree?

Answer 1

Here's a proof of the statement: suppose that $\phi(X) \neq 0 $. That is, suppose that there is some $x \in X$ such that $\phi(x) \neq 0$. For any $r \in \Bbb R$, note that $$ \phi\left(\frac{r}{\phi(x)} x\right) = \frac{r}{\phi(x)}\phi(x) = r $$ It follows that $r \in \phi(X)$. So, if $\phi(X) \neq 0$, then $\phi(X) = \Bbb R$.

Answer 2

The image of a linear map is a subspace. What are the subspaces of $\mathbb{R}$?

You don't even need $\phi$ is bounded, which is what took you astray.