Let X be a normed space and X' its dual space (set of bounded linear functionals on X).
If X not equal to {0} show that X' not equal to {0).
$\displaystyle X\neq\{0\}\Rightarrow\exists x_0\in X$ such that $\displaystyle x_0\neq 0$.
$\displaystyle Y:=\mathbb{C}x_0\subset X$ linear subspace, $\displaystyle \phi:Y\to\mathbb{C},\phi(\lambda x_0):=\lambda,\forall \lambda x_0\in Y$, belongs to $\displaystyle Y^{\,\prime}$.
$\displaystyle p:X\to[0,\infty),p(x):=\|\phi\|\cdot\|x\|,\forall x\in X$ is a seminorm on $\displaystyle X$ with $\displaystyle |\phi(y)|\leqslant p(y),\forall y\in Y$.
By the Hahn-Banach theorem, $\displaystyle \exists f:X\to\mathbb{C}$ linear form such that $\displaystyle f(y)=\phi(y),\forall y\in Y$ and $\displaystyle |f(x)|\leqslant p(x),\forall x\in X$.
$\displaystyle \Rightarrow|f(x)|\leqslant\|\phi\|\cdot\|x\|,\fora ll x\in X\Rightarrow f\in X^{\,\prime}$
and $\displaystyle f(x_0)=f(1\cdot x_0)=\phi(1\cdot x_0)=1\neq 0$.
So we found $\displaystyle f\in X^{\,\prime}$ with $\displaystyle f(x_0)\neq 0\Rightarrow f\neq 0\Longrightarrow X^{\,\prime}\neq 0$.
Well, if you have $\displaystyle X$ a (real or complex) linear space, then $\displaystyle p:X\to[0,\infty)$ is called a seminorm if:
1) $\displaystyle p(x+y)\leqslant p(x)+p(y),\forall x,y\in X$ (i.e. $\displaystyle p$ is sublinear) and
2) $\displaystyle p(\lambda x)=|\lambda|\cdot p(x),\forall x\in X,\forall\lambda$ (in $\displaystyle \mathbb{R}$ or $\displaystyle \mathbb{C}$).
The Hahn-Banach theorem in its complex form (which is a quick corollary of the initial Hahn-Banach theorem) states that if we have $\displaystyle X$ a complex linear space, $\displaystyle p$ a seminorm on $\displaystyle X$, $\displaystyle Y$ a linear subspace of $\displaystyle X$, and $\displaystyle g$ a linear form on $\displaystyle Y$ such that $\displaystyle |g(y)|\leqslant p(y),\forall y\in Y$, then there exists a linear form $\displaystyle f$ on $\displaystyle X$ which is equal to $\displaystyle g$ on $\displaystyle Y$ and which satisfies the inequality $\displaystyle |f(x)|\leqslant p(x),\forall x\in X$.
Let me know if you have any more questions
Thank you again that is very helpful and insightful.
I've got a better idea now but still struggling with one thing.
Its the way you define the mapping of Y in the second line
$\displaystyle
$$\displaystyle Y:=\mathbb{C}x_0\subset X
$
It seems strange to me as \mathbb{C} is the symbol for the set of Complex numbers. Should this be written differently?
Please explain this notation if you dont mind.
I defined $\displaystyle Y$ to be the set $\displaystyle \{\lambda x_0 | \lambda\in\mathbb{C}\}$ which can also be denoted $\displaystyle \mathbb{C}x_0$ and it can easily be shown that is a linear subspace of $\displaystyle X$.
Is that what you were asking?!