Show that the integral of all continuous real-valued functions on the interval [0,1] equal to b $\in$ R is a subspace of $R^{[0, 1]}$ if and only if b=0.
So I am assuming that because both the integral and $R^{[0, 1]}$ are over the same interval, that is significant somehow?
Also I can say the zero function is contained in this set.
I'm not sure how to show this is closed under addition and scalar multiplication, and I don't know why b=0 for the integral to be a subspace of $R^{[0, 1]}$.
Can someone help explain this question to me?
Best Answer
Suppose this was a vector space. Take $f$ and $g$ to be continuous functions on $[0,1]$ for which $\int f dx = b$ and $\int g dx = b$. If this were a vector space then we would have $\int (f+g) dx = b$ as well. However, we know that $\int (f+g) dx = 2b$, thus $2b = b$ and $b=0$.
This demonstrates that the set can only be a vector space if $b$ was zero. Now show that given $b$ is zero, this set constitutes a subspace. That is show that the collection of all continuous functions with zero integral is a subspace.