Proof check of inf f properties.

Question

I have to prove that given f and g are bounded functions on $[a,b]$, then if $c>0$, then $c$ inf $f(x)$ = inf $cf(x)$

My proof goes as follow,

So, we know that $\forall x \in [a,b]$, inf $f(x) \leq f(x)$, and because $c>0$ then, by multiplying by $c$ we get, $$c\cdot\inf\ f(x) \leq cf(x)$$. Now we now that the definition of the infimum of $cf(x)$, $\forall x \in [a,b]$, $$\inf \ cf(x) \leq cf(x)$$

And since for both equations this applies to all $x \in [a,b]$, the only way this would be possible is that both are equal, and hence the proof is done.

Is this correct?

Answer 1

It is not true that $\inf cf$ is the only number $d$ such that $$d\leq cf(x)$$ for all $x\in [a,b]$. Indeed, any number less than $\inf cf$ will also satisfy this inequality. So you cannot conclude that $c\inf f=\inf cf$ just because $d=c\inf f$ satisfies this inequality. To prove $c\inf f=\inf cf$, you must show that $c\inf f$ is the greatest number that satisfies this inequality (since that is the definition of $\inf cf$).

(Also, it's not really correct to write something like $\inf f\leq f$, since $\inf f$ is a number and $f$ is a function. Instead you want $f(x)$ on the right side, since you are evaluating $f$ at some specific $x\in[a,b]$.)

Answer 2

The other direction may deduce on this way :

For all $x\in [a,b]$ and $c>0$ , one has $$f(x)=1\cdot f(x)=(c^{-1}c)\cdot f(x)=c^{-1}\cdot \bigg(cf(x)\bigg)\geq c^{-1}\cdot\bigg(\inf\bigg\{cf(x):x\in[a,b]\bigg\}\bigg)$$

This holds for any $x\in[a,b]$ , so we yield that

$$\inf\big\{f(x) :x\in[a,b]\big\}\ge c^{-1}\cdot\bigg(\inf\bigg\{cf(x):x\in[a,b]\bigg\}\bigg)$$

or $$c\cdot\inf\big\{f(x) :x\in[a,b]\big\}\ge \inf\big\{cf(x):x\in[a,b]\big\}$$

Answer 3

Correct me if wrong:

Let $x\in D:=[a,b].$

Show : $\inf (cf(x))= c \inf (f(x)).$

1)$\inf(f(x)) \le f(x)$.

$c \inf (f(x)) \le cf(x)$, $x \in D.$

Hence $c\inf(f(x))$ is a lower bound of $cf(x),$

implies $c\inf(f(x)) \le \inf(cf(x))$, since the latter is the greatest lower bound(The easier part).

2) Left to show:

$\inf(cf(x)) \le c\inf(f(x)).$

Assume:

$L_1:= \inf (cf(x)) > c\inf (f(x))=:c L_2$.

There is a sequence

$y_n:=f(x_n)$ converging to$ \inf(f(x))=L_2.$

The sequence $c(y_n)$ converges to $cL_2.$

For $\epsilon >0$ there is a $n_0$ such that for $n \ge n_0:$

$|cy_n-cL_2| \lt \epsilon.$

$-\epsilon < cy_n -cL_2< \epsilon$, or

$cL_2 -\epsilon < cy_n<\epsilon +cL_2.$

Choose $\epsilon =(1/2)(L_1-cL_2).$

Then $cy_n < (1/2)(L_1-cL_2) + cL_2=$

$(1/2)L_1 +(1/2)cL_2 < L_1.$

Thus $cy_n = cf(x_n)<L_1= \inf(cf(x)),$

a contradiction.

Hence: $\inf(cf(x)) \le c\inf(f(x)).$

Comments welcome.