I would like to answer this:
Let $f : V \rightarrow \mathbb{R}$ be a linear function with $V$ being a normed vector space (NVS). Prove that $f$ is continuous if and only if $\text{ker} \, f := f^{-1}(\{0\})$ is closed.
However, I believe I've found a counterexample and would like some clarification. Consider the normed vector space of polynomials with the sup norm on $[0,1]$ and the differentiation operator $\frac{d}{dx}$. For this operator:
- The kernel consists of constant polynomials, which correspond to real numbers and is closed in the space of all polynomials.
- The operator seems unbounded since $\frac{\| \frac{d}{dx} x^n \|}{\| x^n \|} = n$, which implies that the operator is not continuous.
Given the above, it seems that we have a closed kernel but a non-continuous operator. Where is the flaw in my argument, or is the exercise statement incorrect?
The derivative map isn't a function from $V \rightarrow \mathbb{R}$ so it doesn't meet the criteria for the question.