I have read that the answer is no, but I am unable to prove it.

Give $C^{\infty}([0,1])$ the metric $$d(f,g) = \sum_{j=0}^{\infty} 2^{-j} \frac{||(f-g)^{(j)}||}{1 + ||(f-g)^{(j)}||}$$ associated to the collection of seminorms $||f^{(j)}||$, the sup-norm of the $j$-th derivative. This makes $C^{\infty}([0,1])$ into a complete metric space.

How do you show that the topological vector space $C^{\infty}([0,1])$ is not a Banach space? That is, there does not exist a norm which induces the same topology.

It does not appear to be enough to show merely that the metric does not arise from a norm, since it is very non-canonical.

## Best Answer

Hints: 1)There is no norm on on $C^{\infty}[0,1]$ which makes the derivative continuous. This is because, $||\frac{d}{dx}(e^{nx})|| \leq c||e^{nx}||$ for all values of $n$ isn't possible.

2) Show that in the above metric the derivative map is continuous.