Is $|f(y_1)-f(y_2)|\leqslant||f||||y_1-y_2||$ true in a Banach space?

Question

I have used during a functional analasys course the following identity:

Consider f as a linear functional on a given infinite-dimensional normed space $Y$, then: $|f(y_1)-f(y_2)|\leqslant||f||||y_1-y_2||$.

If we had admitted that Y was an Hilbert space it is known by Riesz theorem there exists a unique element $u\in Y$ such that $f(x)=\langle x, u\rangle$. Applying Cauchy inequality we get: So $\langle x,u\rangle\leqslant ||u|||x||\implies ||f||=u$

Question: However if we are dealing with a Banach space where the inner product is not defined how would the inequality $|f(y_1)-f(y_2)|\leqslant||f||||y_1-y_2||$ hold?

Thanks in advance!

Answer 1

Let $X$ be a Banach space (more generally a normed space) and $\,f: X\to\mathbb R$ a linear functional.

$f$ is said to be bounded (equivalently continuous) if and only if there exists an $M\ge 0$, such that $$ |\,f(x)|\le M\|x\|,\quad\text{for all $x\in X$.} \tag{1} $$ The least $M$ for which $(1)$ holds is defined to be the norm of $f$, i.e., $$ \|\,f\|=\sup_{x\ne 0}\frac{|\,f(x)|}{\|x\|}. $$ Note that $(1)$ implies that $$ |\,f(x)-f(y)|\le M\|x-y\|, $$ for all $x,y\in X$.