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Prove the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$

Note I asked this question earlier here: Prove the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$, in which I provided a horrendously wrong proof (it's what lack of sleep does). Here I've written a new proof, which I believe should be correct. If it's not please feel free to tear it apart.


Main definition used in proof:

If $S$ is a set, then $F^S$ denotes the set of functions from $S$ to $F$


My Attempted Proof: $$\mathbb{R}^{[0,1]} := \{ f \ | \ f : [0,1] \to \mathbb{R}\}$$

Put $V = \{f \ | \ f: [0,1] \to \mathbb{R} \ \ \text{such that $f$ is continous}\}$

Part 1 :Take $f_0 : [0,1] \to 0$ (i.e. $f_o(x) = 0 \ \ \ \forall x \in [0,1]$). Clearly $f_0$ is continuous and $f_0 \in V$. Now $f_0 + g = g$ for all $g \in V$, and thus $f_0$ is the additive inverse of $V$.

Part 2 : Now take $f,g \in V$.

Now recall for $f, g \in F^S$, the sum $f+g \in F^S$ is the function defined by $(f+g)(x) = f(x) + g(x)$.

Since $V \subset \mathbb{R^{[0,1]}}$ we have $f,g \in \mathbb{R^{[0,1]}}$, and thus the above definition of the addition operation holds, and $f(x) + g(x) = (f+g)(x)$.

And since $f$ and $g$ are both continuous, $f+g$ is also continuous, and combining with what we've shown above we can see that $f+g \in V$. Thus $V$ is closed under addition.

Part 3 :Finally we take $\gamma \in \mathbb{R}$ and $f \in V$.

Recall for $\lambda \in F$ and $f \in F^S$, the product $\lambda f \in F^S$ is the function defined by $(\lambda f)(x) = \lambda f(x)$. So all we have to do is verify that $f \in \mathbb{R^{[0,1]}}$, and scalar multiplication as defined above for $F^S$ (arbitrary function spaces) will hold.

Since $V \subset \mathbb{R^{[0,1]}}$ we have $f \in \mathbb{R^{[0,1]}}$, and $(\gamma f)(x) = \gamma f(x)$

Since $f$ is continuous, $\gamma f$ is also continuous and combining with what we've shown above we can see that $\gamma f \in V$, and thus $V$ is closed under scalar multiplication. $\square$


Is my proof correct and logical/rigorous? And if what I've written is nonsense, please say so.

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Part 1 is o.k. But in Part 2 and 3 you only have considered functions which are constant!

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  • $\begingroup$ I've rewritten the whole of parts 2 and 3, can you take another look if possible? $\endgroup$ – Perturbative Nov 5 '16 at 16:37
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It looks correct logically, but I would argue that the point of the exercise is to prove (in detail) statements such as "if $f$ and $g$ are continuous then $f+g$ is continuous." Although everything you have said is correct, you have seemingly swept the difficulties of the problem under the rug, so to speak.

For example, in Part 2: if we assume that $f, g\in V$ then we wish to show that $f+g\in V$. Indeed, since $f$ and $g$ are continuous then for each $\epsilon>0$ and for each $x \in [0,1]$ there exists a $\delta>0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon/2$ and $|g(x)-g(y)|<\epsilon/2$. From here, you should try to deduce that the function $f+g$ is continuous by showing $|(f+g)(x) - (f+g)(y)|<\epsilon$.

Something similar should be done for Part 3.

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