Let $X$ be a finite dimensional space. I want to prove that all linear functionals are bounded.
MY TRIAL
Let $\dim X=n\geq 1,$ and $\{e_i\}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars $\{\alpha_i\}^{n}_{i=1}$ such that for arbitrary $x\in X,$ \begin{align} x=\sum^{n}_{i=1}\alpha_i e_i.\end{align}
Let $x\in X$ be arbitrary. Then,
\begin{align} |f(x)| &=\left|f\left(\sum^{n}_{i=1}\alpha_i e_i\right)\right|\\&=\left|\sum^{n}_{i=1}\alpha_i f\left(e_i\right)\right|\\&\leq\sum^{n}_{i=1}\left|\alpha_i\right| \left|f\left(e_i\right)\right|\\&\leq \max\limits_{1\leq i\leq n}\left|\alpha_i\right|\sum^{n}_{i=1}\left|f\left(e_i\right)\right|\end{align} Define $\|\cdot\|_0$ as $\|x\|_0=\max\limits_{1\leq i\leq n}\left|\alpha_i\right|$ which is a norm on $X$ [see Prove that $\| \cdot \|_0$ defined by $\| x \|_0=\max\limits_{1\leq i\leq n}|\alpha_i|$ is a norm on $E$.. Hence, \begin{align} |f(x)| \leq \|x\|_0\sum^{n}_{i=1}\left|f\left(e_i\right)\right|\end{align} But All norms defined on a finite dimensional normed linear space are equivalent, so any norm $\| \cdot \|$ defined on $E,$ is equivalent to $\| \cdot \|_0$, i.e. \begin{align} |f(x)| \leq \|x\|\sum^{n}_{i=1}\left|f\left(e_i\right)\right|=K\|x\|,\;\;\text{where}\;K:=\sum^{n}_{i=1}\left|f\left(e_i\right)\right|\geq 0,\end{align} and we are done! Please, I'm I right? Alternative proofs will be highly welcome!
