How to proof $\text{inf}( \lambda A )= \lambda \text{inf} A$

Question

Let $\lambda \in \mathbb R, \lambda \ge 0, \quad \lambda A := \{\lambda x: x \in A\}$

How to proof $\text{inf} \lambda A = \lambda \text{inf} A$?

I have tried to do it with a direct proof like that:

$$\text{inf} \lambda A = \exists c \in \mathbb R, \forall a \in \lambda A : c \le a_{\lambda}$$ $$\equiv \exists c \in \mathbb R, \lambda \in \mathbb R, \forall a \in A: c \le \lambda a $$

Note: I think that I have understood whyt it should be $\text{inf} \lambda A = \lambda \text{inf} A$ intuitively. But unfortunately I am not able to really finish that proof.

I appreciate any help!

Answer 1

If $\lambda = 0$ that is trivial. If it is not, you have $$\lambda \inf A \leq \lambda x, \quad \forall x \in A. (*)$$ Now fix $\varepsilon > 0$. By definition of $\inf A$ you know that you can find $\bar{x} \in A$ such that $$\inf A \leq \bar{x} < \inf A + \varepsilon.$$ Multiply by $\lambda$ (which is positive, so the inequalities do not change): $$\lambda \inf A \leq \lambda \bar{x} < \lambda \inf A + \lambda \varepsilon = \lambda \inf A + \varepsilon' (**)$$ where $\varepsilon'$ is an arbitrary positive number. (*) and (**) give you that $$\lambda \inf A = \inf \lambda A.$$

Answer 2

Hint

Let $\ell=\inf \lambda A$. Prove (using the definition) that $\inf A=\frac{\ell}{\lambda }$.

Answer 3

If $l=\inf A$ then by definition $x\leq l$ and consequently $\lambda x\leq\lambda l$ for each $x\in A$.

This means that $\lambda l$ is an upperbound of $\lambda A:=\{\lambda x\mid x\in A\}$.

Now it remains to prove that $\lambda l$ is the least upperbound of $\lambda A$.

Let us assume now that it is not the least upperbound of $\lambda A$, so that finding a contradiction is enough.

Some $x\in A$ must exist with: $$\lambda l<\lambda x\tag1$$

This cannot be true if $\lambda=0$ and combined with the knowledge that $\lambda\geq0$ we conclude that $\lambda>0$.

Then we can divide both sides of $(1)$ by $\lambda$ and the inequality will stay valid:$$l< x\tag2$$

However $(2)$ contradicts that $l=\inf A$, and we are ready.

Answer 4

assume lambda does not vanish

Then $$\lambda \inf A \leq \lambda x, \quad \forall x \in A. $$

that is $$\lambda \inf A \leq y, \quad \forall y\in \lambda A$$ Hence.
$$\lambda \inf A \leq \inf(\lambda A. )$$ the same reasoning hold if you set $B =\lambda A$ then $A=\lambda^{-1}B$ then we have $$\lambda^{-1} \inf B \leq \inf(\lambda^{-1} B )$$

that is $$\lambda \inf A \geq \inf(\lambda A. )$$