Linear functional is continuous $\implies$ it is bounded Let $f:X \rightarrow \mathbb R$ be a continuous linear functional. Prove $f$ is bounded.

Since $f$ is continuous, $\forall \varepsilon >0$, there exists $\delta >0$ such that $|f(x)-f(y)|=|f(x-y)|=|f(z)|< \varepsilon$ whenever $|x-y|<\delta$. We let $z=x-y$.
Can we just now let $\varepsilon = C \|x\|_X$ for some $C>0$ and then it is bounded?
 A: Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have
$$
1\geq\left|f\left(\frac{\delta x}{\|x\|}\right)\right|=\frac{\delta}{\|x\|}|f(x)|\implies|f(x)|\leq\frac{1}{\delta}\|x\|.
$$
Therefore, $f$ is bounded.
A: Your estimate holds only in a neighborhood of $x$. Hence, you need can not control whether you may choose such a $C$ independent of $x$. You need to rely on the linearity of your operator in the following way:
By continuity of $f$ in $0$, there exists $\delta>0$ such that for all $x\in X$ with $\Vert x \Vert_X <\delta$ holds
$$\vert f(x) - f(0) \vert \leq 1.$$
Using the linearity of $f$ we obtain
$$\vert f(x) \vert = \vert f(x) - 0 \vert =\vert f(x) - f(0) \vert \leq 1.$$
For $y\in X$, $y\neq0$ we get again by linearity of $f$
$$ \vert f(y) \vert  = \frac{\Vert y \Vert_X}{\delta} \cdot \underbrace{\Vert f\left(\frac{\delta y}{\Vert y \Vert_X} \right) \Vert_X}_{\leq 1}  \leq \frac{1}{\delta} \Vert y \Vert_X.$$
This is presicely the definition of bounded operator.
