Suppose $x\in X$ with $x\neq 0$. Then $y = x/\|x\|$ satisfies $\|y\| = 1$, and hence $|f(y)|\leq \|f\|$. But $f(y) = f(x/\|x\|) = f(x)/\|x\|$, so $|f(x)|/\|x\|\leq \|f\|$. This proves the inequality $|f(x)|\leq \|f\|\|x\|$ when $x\neq 0$. The inequality is trivial with $x = 0$.
There are (at least) two somewhat conflicting definitions of a dual space.
In functional analysis, we start with a topological vector space $V$, usually over real or complex numbers, and then the (continuous) dual is the space of all continuous functionals. For normed (in particular, Banach) spaces, a functional is continuous if and only if it is bounded.
In abstract algebra, we deal with vector spaces over arbitrary fields. Frequently, the vector spaces are finite-dimensional, in which case the dual actually coincides with algebraic dual, but for infinite dimensional spaces (over real or complex numbers), the algebraic dual is usually larger by far than the continuous dual*.
More abstractly, the difference of definition is a matter of perspective. In terms of category theory, you can think of the dual of a $K$-vector space $V$ as the space $\operatorname{Hom}(V,K)$, i.e. the set of morphisms from $V$ to the base field $K$ in your category. If the category is that of vector spaces, you obtain the algebraic dual. If the category is that of topological vector spaces, you obtain the continuous dual.
*${}$a pure vector space can be regarded as a discrete topological vector space. In this case, it is fairly easy to see that the continuous dual and the algebraic dual coincide.
Best Answer
By linearity, $\frac{|f(u)|}{\|u\|} = |f(\frac{u}{\|u\|})|$ so that's one direction, that is (2) (the 2nd def) is $\leq (1)$. The other direction follows from the fact that if $\|u\|\leq 1$, then $|f(u)| \leq \frac{|f(u)|}{\|u\|}$