1. ## Dual Space Problem

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).

2. $X\neq\{0\}\Rightarrow\exists x_0\in X$ such that $x_0\neq 0$.
$Y:=\mathbb{C}x_0\subset X$ linear subspace, $\phi:Y\to\mathbb{C},\phi(\lambda x_0):=\lambda,\forall \lambda x_0\in Y$, belongs to $Y^{\,\prime}$.
$p:X\to[0,\infty),p(x):=\|\phi\|\cdot\|x\|,\forall x\in X$ is a seminorm on $X$ with $|\phi(y)|\leqslant p(y),\forall y\in Y$.
By the Hahn-Banach theorem, $\exists f:X\to\mathbb{C}$ linear form such that $f(y)=\phi(y),\forall y\in Y$ and $|f(x)|\leqslant p(x),\forall x\in X$.
$\Rightarrow|f(x)|\leqslant\|\phi\|\cdot\|x\|,\fora ll x\in X\Rightarrow f\in X^{\,\prime}$
and $f(x_0)=f(1\cdot x_0)=\phi(1\cdot x_0)=1\neq 0$.
So we found $f\in X^{\,\prime}$ with $f(x_0)\neq 0\Rightarrow f\neq 0\Longrightarrow X^{\,\prime}\neq 0$.

3. ## Great

Great AMI thank you for your help.
I am still working through it.
I dont have much experience with semi-norm. What is this exactly?
I understand Hahn-Banach Theorem.

Much appreciated for your time and effort.

Originally Posted by AMI
$X\neq\{0\}\Rightarrow\exists x_0\in X$ such that $x_0\neq 0$.
$Y:=\mathbb{C}x_0\subset X$ linear subspace, $\phi:Y\to\mathbb{C},\phi(\lambda x_0):=\lambda,\forall \lambda x_0\in Y$, belongs to $Y^{\,\prime}$.
$p:X\to[0,\infty),p(x):=\|\phi\|\cdot\|x\|,\forall x\in X$ is a seminorm on $X$ with $|\phi(y)|\leqslant p(y),\forall y\in Y$.
By the Hahn-Banach theorem, $\exists f:X\to\mathbb{C}$ linear form such that $f(y)=\phi(y),\forall y\in Y$ and $|f(x)|\leqslant p(x),\forall x\in X$.
$\Rightarrow|f(x)|\leqslant\|\phi\|\cdot\|x\|,\fora ll x\in X\Rightarrow f\in X^{\,\prime}$
and $f(x_0)=f(1\cdot x_0)=\phi(1\cdot x_0)=1\neq 0$.
So we found $f\in X^{\,\prime}$ with $f(x_0)\neq 0\Rightarrow f\neq 0\Longrightarrow X^{\,\prime}\neq 0$.

4. Well, if you have $X$ a (real or complex) linear space, then $p:X\to[0,\infty)$ is called a seminorm if:
1) $p(x+y)\leqslant p(x)+p(y),\forall x,y\in X$ (i.e. $p$ is sublinear) and
2) $p(\lambda x)=|\lambda|\cdot p(x),\forall x\in X,\forall\lambda$ (in $\mathbb{R}$ or $\mathbb{C}$).

The Hahn-Banach theorem in its complex form (which is a quick corollary of the initial Hahn-Banach theorem) states that if we have $X$ a complex linear space, $p$ a seminorm on $X$, $Y$ a linear subspace of $X$, and $g$ a linear form on $Y$ such that $|g(y)|\leqslant p(y),\forall y\in Y$, then there exists a linear form $f$ on $X$ which is equal to $g$ on $Y$ and which satisfies the inequality $|f(x)|\leqslant p(x),\forall x\in X$.

Let me know if you have any more questions

5. ## almost there

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

$
$
$Y:=\mathbb{C}x_0\subset X
" alt="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.

Originally Posted by AMI
Well, if you have $X$ a (real or complex) linear space, then $p:X\to[0,\infty)$ is called a seminorm if:
1) $p(x+y)\leqslant p(x)+p(y),\forall x,y\in X$ (i.e. $p$ is sublinear) and
2) $p(\lambda x)=|\lambda|\cdot p(x),\forall x\in X,\forall\lambda$ (in $\mathbb{R}$ or $\mathbb{C}$).

The Hahn-Banach theorem in its complex form (which is a quick corollary of the initial Hahn-Banach theorem) states that if we have $X$ a complex linear space, $p$ a seminorm on $X$, $Y$ a linear subspace of $X$, and $g$ a linear form on $Y$ such that $|g(y)|\leqslant p(y),\forall y\in Y$, then there exists a linear form $f$ on $X$ which is equal to $g$ on $Y$ and which satisfies the inequality $|f(x)|\leqslant p(x),\forall x\in X$.

Let me know if you have any more questions

6. I defined $Y$ to be the set $\{\lambda x_0 | \lambda\in\mathbb{C}\}$ which can also be denoted $\mathbb{C}x_0$ and it can easily be shown that is a linear subspace of $X$.
Is that what you were asking?!