Proving $\sup{|\lambda \cdot u(x)|} = |\lambda| \sup{|u(x)|}, \forall \lambda \in R$

Question

I was proving that a certain function define a norm in $C([0, 2])$, and while verifying $||\lambda \cdot u||=|\lambda| \cdot ||u||$, I needed to demonstrate: $$\underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} = |\lambda|\underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|},\,\forall \lambda \in \Bbb R$$

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My attempt:

1. $\forall \lambda \in \Bbb R, \; \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} = \max\left[\,\left|\underset{0\le x \le1}{\inf}{(\lambda \cdot \operatorname{u}(x))}\right|, \, \underset{0\le x \le1}{\sup}{(\lambda \cdot \operatorname{u}(x))}\,\right]$

2. $\lambda \ge 0 \rightarrow \begin{cases} \underset{0\le x \le1}{\inf}{(\lambda \cdot \operatorname{u}(x))} = \lambda \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))} \\[2ex] \underset{0\le x \le1}{\sup}{(\lambda \cdot \operatorname{u}(x))} = \lambda \cdot \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))} \end{cases}$

3. $\lambda \le 0 \rightarrow \begin{cases} \underset{0\le x \le1}{\inf}{(\lambda \cdot \operatorname{u}(x))} = \lambda \cdot \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))} \\[2ex] \underset{0\le x \le1}{\sup}{(\lambda \cdot \operatorname{u}(x))} = \lambda \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))} \end{cases}$

4. If $\lambda \ge 0$, then:
   \begin{align} \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} & = \max\left[\,\left|\lambda \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\right|, \, \lambda \cdot \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\,\right] \\ & = \max\left[\, \lambda \cdot \left|\underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\right|, \, \lambda \cdot \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\,\right] \\ & = \lambda \cdot \max\left[\, \left|\underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\right|, \, \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\,\right] \\ & = \lambda \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|} \\ & = |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}\ \end{align}

5. If $\lambda \le 0$, then:
   \begin{align} \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} & = \max\left[\,\left|\lambda \cdot \underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\right|, \, \lambda \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\,\right] \\ & = \max\left[\, |\lambda| \cdot \left|\underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\right|, \, \lambda \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\,\right] \\ & = \max\left[\, |\lambda| \cdot \left|\underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\right|, \, -|\lambda| \cdot \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\,\right] \\ & = |\lambda| \cdot \max\left[\,\left|\underset{0\le x \le1}{\sup}{(\operatorname{u}(x))}\right|, \, - \underset{0\le x \le1}{\inf}{(\operatorname{u}(x))}\,\right] \\ & = \text{I'm stuck here.} \end{align}

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• Is my demonstration correct until here?
• Can somebody help me to continue?
• Or, can someone suggest me another way?

Answer 1

Easier way: For any $x$ we have $ |\lambda u(x)|=|\lambda| |u(x)| \leq |\lambda| \sup_{0 \leq x \leq 1} |u(x)|$ and this implies that LHS $\leq$ RHS. Similarly $ |\lambda | | u(x)|=|\lambda u(x)| \leq \sup_{0 \leq x \leq 1} |\lambda u(x)|$ and this implies that LHS $\leq$ RHS.

Answer 2

1. $\forall \lambda \in \Bbb R$, I define the set: $A = \{y\in \Bbb R\mid y=|\lambda \cdot \operatorname{u}(x)|, \, x \in [0,1]\}$.
2. $\forall x \in [0,1]$, I get: $|\lambda \cdot \operatorname{u}(x)| = |\lambda| \cdot |\operatorname{u}(x)|\le |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}$.
3. So, $\left(|\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}\right)$ is an upper bound for $A$.
4. By definition, we have: $\sup(A) = \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} \le |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}$

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1. $\forall \lambda \in \Bbb R$, I define the set: $B = \{y\in \Bbb R\mid y=|\lambda| \cdot |\operatorname{u}(x)|, \, x \in [0,1]\}$.
2. $\forall x \in [0,1]$, I get: $|\lambda| \cdot |\operatorname{u}(x)| = |\lambda \cdot \operatorname{u}(x)|\le \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|}$.
3. So, $\left(\underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|}\right)$ is an upper bound for $B$.
4. By definition, we have: $\sup(B) = \underset{0\le x \le1}{\sup}{\left(|\lambda| \cdot |\operatorname{u}(x)|\right)} = |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|} \le \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|}$

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Now we have:

1. $\underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} \le |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}$
2. $|\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|} \le \underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|}$

From that, it follows: $\left(\underset{0\le x \le1}{\sup}{|\lambda \cdot \operatorname{u}(x)|} = |\lambda| \cdot \underset{0\le x \le1}{\sup}{|\operatorname{u}(x)|}\right)$