If $f$ is a Sobolev $H^1$-function and vanishes everywhere except for a point, is it constantly zero?

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Let $H^1(\mathbb{R}^3)$ be the usual $L^2$-based Sobolev space. If $f\in H^1(\mathbb{R}^3)$ satisfies $f=0$ in $\mathbb{R}^3\setminus\{0\}$, can one conclude that $f=0$ in $H^1(\mathbb{R}^3)$?

I think this is true because the condition implies $f$ can only be supported at $\{0\}$, so it must be a linear combination of derivatives of the delta distribution at $\{0\}$, but none of these distributions are in $H^1(\mathbb{R}^3)$. Is this argument correct?