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?
 A: 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 $\sup(A_i+B_i)\leq \sup(A_i)+\sup(B_i)$. 
A: 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).
$$\|\lambda \phi\| = \sup_{x: \|x\| \le 1} |\lambda \phi(x)| = |\lambda| \sup_{x: \|x\|\le 1} |\phi(x)| = |\lambda| \|\phi\| $$
