Show that $||x||_{\infty}\leq||x||_2$ Equivalent norms on $\Bbb R^d$:
Show that $||x||_{\infty}\leq||x||_2$. When is there equality?
My attempt:
Assume $x$ is the zero vector. Then $||x||_{\infty}=0=||x||_2$
Assume $x=\sum^d_{j=1}x_je_j$ where $e=(e_1,...,e_d)$ is the canonical basis for $\Bbb R^d$, and where $x_j=0$ for all $j$ except $1$. Then $||x||_{\infty}=1=||x||_2$
Now assume $x$ os none of the above. Then since $|y|=\sqrt{y^2}$ and for any $t\in \Bbb R$ $|y| < \sqrt{y^2+t^2}$. Hence $||x||_{\infty}<||x||_2$
Would this be correct?
 A: Let 
$$
\mathbf{x} \colon= \left( x_1, \ldots, x_d \right)
$$
be any point in $\mathbb{R}^d$. Then by definition
$$
\lVert \mathbf{x} \rVert_\infty \colon= \max_{ i \in \{1, \ldots, d \} } \left\lvert x_i \right\rvert.
$$
So we can conclude that
$$
\lVert \mathbf{x} \rVert_\infty = \left\lvert x_k \right\rvert
$$
for some $k \in \{ 1, \ldots, d \}$, and since 
$$
\left\lvert x_i \right\rvert \geq 0 
$$
for each $i \in \{ 1, \ldots, d \}$, therefore we obtain 
$$
\begin{align}
\lVert \mathbf{x} \rVert_\infty &= \left\lvert x_k \right\rvert \\
&= \sqrt{ \left( \left\lvert x_k \right\rvert \right)^2 } \\
&\leq \sqrt{ \sum_{i = 1}^d \left( \left\lvert x_i \right\rvert \right)^2 } \\
&= \lVert \mathbf{x} \rVert_2.
\end{align}
$$
And, 
$$
\lVert \mathbf{x} \rVert_\infty = \lVert \mathbf{x} \rVert_2
$$
if and only if
$$
\sqrt{ \sum_{i = 1}^d \left( \left\lvert x_i \right\rvert \right)^2 } = \left\lvert x_k \right\rvert 
$$
for some $k \in \{ 1, \ldots, d \}$, and the last equality holds if and only if
$$
 \sum_{i = 1}^d \left( \left\lvert x_i \right\rvert \right)^2 = \left( \left\lvert x_k \right\rvert \right)^2 ,
$$
which holds if and only if
$$
\left\lvert x_i \right\rvert = 0,
$$
and hence 
$$
x_i = 0
$$
for all $i \in \{ 1, \ldots, d \}$ except possibly for one value $k \in \{ 1, \ldots, d \}$.
Hope this helps.
