Let $d \in \mathbb N$, and $\mathbb S^{d-1}:=\{x\in \mathbb R^{d}: \vert\vert x \vert\vert_{2}=1\}$ is the unit sphere with regards to $\vert\vert \cdot \vert\vert$ in $\mathbb R^{d}$:
Find the following:
(a) $\sup\{\vert\vert\cdot\vert\vert_{\infty}:x\in \mathbb S^{d}\}$
(b) $\sup\{\vert\vert\cdot\vert\vert_{1}:x\in \mathbb S^{1}\}$
So far:
(a) Since $x\in \mathbb S^{d}$, we're looking at $x:=(x_{1},...,x_{d+1})^{T}$ and the first $x$ that comes to mind such that $(\sum^{d+1}_{k=1}|x_{k}|^{2})^{\frac{1}{2}}=1$ and such that we maximize a component is $x_{k}=1$ for only one $k \in \{1,...,d\}$ with the rest $x_{j}=0$. Is this correct, or do I need better justification?
(b) Since $x \in \mathbb S^{1}$ we're looking at vector $x:=(x_{1},x_{2})$ such that $(\sum^{2}_{k=1}|x_{k}|^{2})^{\frac{1}{2}}=1$.
In essence, we are trying to maximize $|x_{1}|+|x_{2}|$, such that $(\sum^{2}_{k=1}|x_{k}|^{2})^{\frac{1}{2}}=1$ is still satisified. Looking at
$(\sum^{2}_{k=1}|x_{k}|^{2})^{\frac{1}{2}}=1$,
it follows that $|x_{1}|^{2}+|x_{2}|^{2}=1$ and since we are in $\mathbb R$, it follows: $x_{1}^{2}+x_{2}^{2}=1$
I am thinking of using partial derivatives, however, we've jsut started and I'm unsure whether I'm going down the right path.
a) You are correct. Note the following: $\forall j$, $\lvert x_j\rvert = \sqrt{x_j^2} \leq \left( \sum^{d+1} x_i^2 \right) ^{1/2} = 1$
b) Since we are looking to maximize $\lvert x_1\rvert + \lvert x_2\rvert$ on the unit circle, we can assume $x_1,x_2 > 0$ (ie. our solution is in the top-right quadrant). We can denote $x_1=\cos\theta$ and $x_2=\sin\theta$. Maximum of $\cos\theta+\sin\theta$ is achieved when its derivative is zero (ie. $-\sin\theta+\cos\theta=0$ or $\cos\theta=\sin\theta$). This happens when $\theta=\frac{\pi}{4}$. Thus, $x_1=x_2=\frac{1}{\sqrt{2}}$