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My question is in relation to : Approximation by smooth function while preserving the zero set

Let $\mathbb{T}$ denote the unit circle. Given $f \in \mathcal{C}^1(\mathbb{T})$, can we approximate $f$ by smooth functions having the same zero set ? i.e. for $\varepsilon >0$, can we find $g \in \mathcal{C}^\infty(\mathbb{T})$ such that

  1. f(x) = 0 if and only if $g(x) = 0$;
  2. $\|f-g\|_{\mathcal{C}^1(\mathbb{T})}< \varepsilon$.

Here $\|h\|_{\mathcal{C}^1(\mathbb{T})} := \|h\|_\infty + \|h'\|_\infty = \sup\limits_{z \in \mathbb{T}}|f(z)| + \sup\limits_{z \in \mathbb{T}}|f'(z)|$.

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    If $f$ has finitely many zeros, yes, for example let $g_1(x)=\int_0^{2\pi} f(y)h(x-y)dy$ and $g(x) = g_1(\phi(x))$ where $h,\phi$ are smooth and $\phi$ moves the zeros to the right place. Otherwise, not sure what happens at an accumulation point of the zeros – 2017-02-07

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