Is the space $\{f\in C[0,1]\mid \int_0^1f\neq 0\}$ dense in $C[0,1]$ with sup-norm topology.
I think yes, because it is the inverse image of the set $\mathbb{R} \setminus \{0\}$
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@Gribouillis thanks, the answer by Jorge is also great. – vidyarthi Jan 01 '17 at 07:28
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In fact my comment was wrong. A correct related argument would be that your set is the complement of a closed hyperplane. This is always dense in any Banach space, but a direct proof such as Jorge's is probably better. – Gribouillis Jan 01 '17 at 07:33
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Call that set $A$, we shall prove it is a dense set.
Pick $f\not \in A$, in other words $f$ such that $\int\limits_{0}^1 f=0$, and pick $\epsilon >0$.
Notice that the function $g(x)=f(x)+\epsilon/2$ is in $A$ and its distance from $f$ is $\epsilon/2$, so the set is in fact dense.
Asinomás
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1nope, the continuous functions with positive integral form an open subset, and so do the ones with negative integral. – Asinomás Jan 01 '17 at 07:10
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