# Showing dual space of a normed space is a normed space

Let $X$ be a Normed Linear Space. The dual space $X^*$ of $X$ is the set of all bounded linear functionals on $X$. It is a normed linear space with the norm $\Vert\varphi\Vert$. (according to notes).

I can show the triangle inequality is true.

Trying to show $\Vert\varphi\Vert = 0$ iff $\varphi = 0$, I get:

$\varphi = 0$ implies $\Vert\varphi\Vert = 0$. (OK).

$\Vert\varphi\Vert = 0$ implies $\varphi (x) = 0$ for $\Vert\varphi\Vert\leq 1$. But I don't see how this implies $\varphi = 0$.

Also, $\Vert\lambda\varphi(x)\Vert = \vert \lambda \vert \sup\{\vert\varphi(x)\vert \colon \vert\lambda \vert \Vert x \Vert\leq 1\} \neq \vert\lambda\vert \Vert\varphi(x)\Vert$

Where have I gone wrong?

-
Sorry... could you use some terminology instead of persona abbreviations? What is "NLS"? Searching for "NLS math" produces "Non-linear Schrodinger" as the most popular hit. Is it really so hard to type out full words? – Arturo Magidin May 6 '12 at 20:18
Normed Linear space? – user21436 May 6 '12 at 20:26
NLS = Normed linear space. Apologies... I will write out in full in future (even though I find it tedious)... David - All I can say is that |phi| = 0 for all ||x|| <= 1. I can't say anything for ||x|| >= 1. Also, if you are thinking on the lines of ||phi(x)|| = ||phi|| ||x|| (because phi is bounded), then you have skipped a few lines ahead in my notes... otherwise I can't see what to do. – Adam Rubinson May 6 '12 at 20:30
Hint: for general non-zero $x$, we have $x={\Vert x \Vert x\over\Vert x\Vert}$, and $\varphi\bigl({\Vert x\Vert x\over \Vert x\Vert} \bigr)= \Vert x\Vert {\varphi({\underbrace{x\over \Vert x\Vert }_{ {\text{ norm 1}}} }) }$. – David Mitra May 6 '12 at 20:40
There is no reason to have $|\lambda|\lVert x\rVert \leq 1$ in your computation of $\lVert\lambda\varphi\rVert$; why did you put that into the formula? The definition says to look at all $x$ with $x\leq 1$. – Arturo Magidin May 6 '12 at 20:57

By definition $\lVert\varphi\rVert = \sup\{|\varphi(x)|\mid \lVert x\rVert \leq 1\}$.

So, if $\lVert \varphi\rVert = 0$, $|\varphi(x)|=0$ for all $x$ with $\lVert x\rVert =1$. Let $x\neq 0$. Then $w=\frac{1}{\lVert x\rVert}x$ has norm $1$, so $\varphi(w) = 0$. But $\varphi(w) = \frac{1}{\lVert x\rVert}\varphi(x)$, hence $\varphi(x)=0$. Thus, $\varphi(x)=0$ for all $x$.

Now, \begin{align*} \lVert \lambda \varphi\rVert &= \sup\{ |\lambda\varphi(x)|\mid \lVert x\rVert \leq1\}\\ &= \sup\{ |\lambda| |\varphi(x)|\mid \lVert x\rVert \leq 1\}\\ &= |\lambda|\sup\{|\varphi(x)|\mid \lVert x\rVert \leq 1\}\\ &= |\lambda|\lVert \varphi\rVert. \end{align*} (I think you got confused on this by not writing it out carefully).

Finally, \begin{align*}\lVert \varphi+\psi\rVert &= \sup\{ |(\varphi+\psi)(x)|\mid \lVert x\rVert \leq1\}\\ &= \sup\{ |\varphi(x)+\psi(x)|\mid \lVert x\rVert \leq 1\}. \end{align*}

Now use the triangle inequality to get $|\varphi(x)+\psi(x)|\leq |\varphi(x)|+|\psi(x)|$; use the fact that if $\{A_i\}$ and $\{B_i\}$ are two families and $A_i\leq B_i$ for each $i$, then $\sup\{A_i\}\leq\sup\{B_i\}$; and then use the fact that $\sup(A_i+B_i)\leq \sup(A_i)+\sup(B_i)$.

-
Yes. I was thinking something that isn't true – Adam Rubinson May 6 '12 at 20:58

If $\phi(x) = 0$ for all $x$ with $\|x\|\le 1$, then since $\phi(tx) = t \phi(x)$ we get $\phi(x)=0$ for all $x$. So $\phi = 0$ (i.e. these are the same linear functionals).

$$\|\lambda \phi\| = \sup_{x: \|x\| \le 1} |\lambda \phi(x)| = |\lambda| \sup_{x: \|x\|\le 1} |\phi(x)| = |\lambda| \|\phi\|$$

-
Thanks for all your responses. – Adam Rubinson May 6 '12 at 21:33