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Consider the sequence of functions \begin{align*} f_n(x)=\begin{cases} 0 & x \leq -\tfrac{1}{n} \\[0.5em] nx+1 & x \in (-\tfrac{1}{n},0) \\[0.5em] 1 & x\geq 0 \end{cases} \end{align*} Show by the definition of a distribution that $f_n\rightarrow H$ in the sense of distributions where $H$ is the Heaviside function.

I feel like it would be easier to show that $f_n-H\rightarrow 0$ in the sense of distributions, but I'm not sure how to go from here.

Any tips would be appreciated!

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We have to show that $\int_0^{\infty} \phi (x) dx+\int_{-1/n}^{0} (1+nx) \phi (x)dx \to \int_0^{\infty} \phi (x) dx$ for every test function $\phi$. So it suffices to show that $\int_{-1/n}^{0} (1+nx) \phi (x)dx \to 0$. Use the fact that $|1+nx| \leq 1$ and $\phi$ is integrable. So DCT finishes the proof.

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  • $\begingroup$ could you elaborate further? also, i'm assuming DCT is the Distribution Convergence Theorem $\endgroup$
    – user481532
    Oct 14, 2019 at 23:59
  • $\begingroup$ $|\int_{-1/n}^{0} (1+nx) \phi (x)dx| \leq \int_{-1/n}^{0} |\phi (x)| dx $. For any integrable function $\phi$, $\int_A \phi (x)| dx \to 0$ as $mu (A) \to 0$. @nickoba $\endgroup$
    – Kavi Rama Murthy
    Oct 15, 2019 at 0:02

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