Are Continuous and Differentiable Functions on [0,1] Closed Under Operations?

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If G=the set of all continuous complex-valued functions on the interval [0,1] and f,g \in G then is \displaystyle f(x) \int^1_0 g(t) \; dt - g(x) \int^1_0 f(t)\;dt in G?

If H=the set of all differentiable complex-valued functions on the interval [0,1] and f,g \in H then is fg' - gf' in H?
 
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\int_0^1 f(x) dx and \int_0^1 g(x)dx are numbers so, knowing that f and g are continuous, what can you say about Af+ Bg for constants B and G?

The derivative of f'g+ fg' is f''g+ 2f'g'+ fg''. Knowing that f and g are differentiable what can you say about f'' and g''?
 
HallsofIvy said:
\int_0^1 f(x) dx and \int_0^1 g(x)dx are numbers so, knowing that f and g are continuous, what can you say about Af+ Bg for constants B and G?

The derivative of f'g+ fg' is f''g+ 2f'g'+ fg''. Knowing that f and g are differentiable what can you say about f'' and g''?

Got it. They're both still in the sets.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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