Proving $|f(x)|$ $\le \| x\|\| f\|$ , where f is a continuous linear functional I am self studying Functional Analysis and came across this proposition.
Suppose, $f$ is a continuous linear functional defined on a Normed Linear space $V$, then
$$|f(x)| \le \| x\|\| f\|$$
I don't get how this inequality is derived, I know $f : V \to \mathbb{R}$ is a continuous  linear map
So, we have $\| f(x)\| \le \| f\|\| x\|$
Now, are we assuming the norm on $\mathbb{R}$ to be the usual $l^1$ norm in this case?
 A: Yes I think you're right that we're assuming the standard norm on R. You could also check the definition of the norm of a linear functional given in your text. It may explicitly use the absolute value.
A: Adding to the answer of Physor, which proved the inequality for all norms in $\mathbb{R}$, in particular for the usual one. I added a answer for your second question:
Yes, he is assuming that the norm in $\mathbb{R}$ is the usual one. The
reason for this is that in $\mathbb{R}$, since it is a finite dimensional vector space, all norms are equivalent, and then all of these norms have the form $\lambda |x|$ where $\lambda>0$ is a fixed real number. And so the inequality that are you considering changes only by a multiplication by constant when changing the norms in $\mathbb{R}$, which does not matter in terms of convergence. So, yes, there is no loss in generality in consider only the usual norm in this case.
A: This is a direct proof:
Let $f:V \to W$ be a continuous linear map between normed spaces $V,W$, i.e.
$$
\forall \varepsilon > 0 :\exists \delta>0 :\forall x \in V:(\|x\| < \delta \implies\|f(x)\| < \varepsilon).
$$
It follows that for every preassigned $\varepsilon$ and as long as $ 0 <\|x\| < \delta(\varepsilon)$ we have
$$
\|f(x)\| < \varepsilon \implies \frac{\|f(x)\|}{\|x\|} < \frac{\varepsilon}{\|x\|}.
$$
Now we can take the infimum on RHS of the last inequality
$$
\frac{\|f(x)\|}{\|x\|} \le \inf_{0 < \|x\| < \delta(\varepsilon)} \left\{ \frac{\varepsilon}{\|x\|} \right\} = \frac{\varepsilon}{\delta(\varepsilon)}=:\alpha(\varepsilon) < \infty,
$$
since $\delta(\varepsilon) > 0$. Thus we proved, since the RHS of the last inequality doesn't depend on $\|x\|$,  that
$$
\exists\alpha \ge 0 :\forall x \in V: \|f(x)\| \le\alpha \|x\|
$$
which is boundedness of $f$. The infimum of all these $\alpha(\varepsilon)$ is the norm $\|f\|$ of $f$.
