# Show the usual Schwartz semi-norm is a norm on the Schwartz space

Let $f \in C^\infty(\mathbb R)$. Define the semi-norm $$\|f\|_{a,b}=\sup_{x \in \mathbb R} |x^af^{(b)}(x)|$$ where $a,b \in \mathbb Z_+$, and $f^{(b)}$ is the $b$-th derivative of $f$.
Show $\|\cdot\|_{a,b}$ is a norm on the Schwartz space $S(\mathbb R)$.

I don't see how to proove this direction of the nonnegativity requirement for the norm $$\|f\|_{a,b} =0 \text{ implies } f=0.$$

If one of the derivative is zero, how can I infer that the original function is also zero?

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## 1 Answer

Let $f$ such that $\lVert f\rVert_{a,b}=0$. Then for each $x\neq 0$, we have $f^{(b)}(x)=0$, and as $f^{(b)}$ is continuous, $f^{(b)}(x)=0$ for all $x\in\Bbb R$. This implies that $f$ is a polynomial. As $f$ is assumed to be in the Schwartz space, $f$ vanishes at infinity, hence $f\equiv 0$.

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Very good proof. Thank you. – Nicolas Essis-Breton Nov 2 '12 at 18:04
Actually, the only zero elements on the Schwartz space such that $||f||_{a,b}=0$ is the null function. For deducing that $f$ is a polynomial, I do not require it to be in the Schwarz space, I just use the fact that $f$ is smooth. – Davide Giraudo May 22 at 20:55