Let $F$ be the vector space of all functions $R \to R$. Let $U_2$ be the subset of F formed by all functions bounded by above (a function $f: R \to R$ is bounded from above if there exists $M ∈ R$ such that $$ f(x) ≤ M, \ \ \text{for all} \ x ∈ R. $$ Show that $U_2$ is not a subspace of $F$. Question
I am confused with the process to check if $U_2$ is not closed under scalar multification.
Here is what I did. Take any $u ∈ U_2$, any scalar $c ∈ R$ $c*f(x) ≤ c*M$; for all $x \in R$
When c < 0, $$c*f(x) ≥ c*M$$ I thought when $c < 0$, $f$ is not in $U_2$ because $c*f(x) ≥ c*M$ is not the form of $f(x) ≤ M$.
Let's consider when $f(x) = -x^2$, then f(x) ≤ 0 and -f(x) ≥ 0.
However, "The fact that $-f(x) = x^2 ≥ 0$ doesn't mean that $-f$ is not in $U_2$"
This is what I got as my feedback.
"The function $f(x) = -x^2$ is in $U_2$ because it is bounded above : there exists a real number $M$ (here $M=0$ works) such that for all $x$, $f(x) ≤ M.$ But you can't find any $M'$ such that $-f(x) ≤ M'$ for all x, so $-f$ is not in $U_2$."
This is the solution from my prof. Don't I need to change that inequality sign when I use a scalar?
"This is not because $-f(x) >= -M$ that $-f$ is not in $U_2$. The constant function $f(x)=1$ is in $U_2$, and $-f(x)=-1$ is again in $U_2$."
According to this, constant $f(x)$ always works under the condition. Then how am I supposed to prove some $f(x)$ doesn't work. I would like to make this generalized without some specific example.
Do I need to find an example all the time?
Please give me some insight on this question. I am dying to understand this.
Thank you!