# Which of the following are subspaces of V, a vector space over the reals?

Let V be the set of functions f: [0,1] -> Reals; it is a vector space over the reals. Of the following, which are subspaces of V? (a) {f such that f is bounded} (b) {f such that abs(f) is bounded by 1} (c) {f such that f is increasing} (d) {f such that f(0)=0}

I know the criteria a subset must follow in order for it to be a subspace (contain 0 vector, remain closed under addition/scalar mult.) but I don't know how to apply them here.

For example, for (a), is it not a subspace, because it is bounded, and such if we chose 2 vectors that, when added, exceeded the bound of f, then it wouldn't hold (?) And then I think the same would be true for (b), but I'm not sure. Please help

• a) yes $|f+ag|\leq |f|+|a||g|$, b)No, C)no for $-f$, d)yes $f+ag(0)=f(0)+ag(0)=0+0=0$ – R.N Nov 11 '15 at 22:49

## 2 Answers

Your reasoning for $(a)$ doesn't quite work, although it does apply for $(b)$: the set of functions in $(a)$ is the set of functions which have any bound. So, if $f$ is bounded by $a$ and $g$ is bounded by $b$, then $f+g$ is bounded by $a+b$; the bound is allowed to change. For $(b)$, the bound is fixed (specifically, 1), so you are correct: the answer for $(b)$ is "no."

Now on to $(c)$ and $(d)$. Both sets of functions are closed under addition. So we need to ask about the other axioms:

• Contain the "zero vector" (what is the zero vector for $V$?), and

• closed under scalar multiplication.

Do you see how to proceed with these?

a) and d) are subspaces of $V$ but b) and c) are not.

To prove this proceed as you said. I will just prove the first two cases, the other two can be proven in the same way.

a) Let's denote that set as $V_a$. Then the function $0$ belongs to $V_a$ because it is bounded. Then let's consider $f,g \in V_a$ such that $|f(x)| \le M$ and $|g(x)| \le M'$ $\forall x \in \mathbb{R}$. So the function $(|\alpha f + \beta g|)(x) \le |\alpha| M + |\beta| M'$ is bounded, so $V_a$ is a subspace of $V$.

In the case of b) consider $f(x) = 1 ~ \forall x \in \mathbb{R}$. Then $f \in V_b$ but $2 f \notin V_b$, so it is not a subspace.