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I know that convergence in $D'(\Omega)$ (The space of distributions) is just the convergence in $\mathbb{R}$ for every test function $\phi$. But I am not sure about something like " Norm convergence implies convergence is distributions" , what about $L^{p}$ convergence implies convergence in distributions and $L_{loc}^{2}$ etc.A sequence $T_{n}$ of distributions convergence to another distribution $T$ if we have $\lim_{n \rightarrow \infty} \langle T_{n},\phi \rangle = \langle T, \phi \rangle$ for ever test function $\phi$. I appologize if it seems something obvious. Any discussion will be highly appreciated.

Mathslover shah
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To every function $u\in L^p(\Omega)$ you can associate a distribution $T_u$ defined by $$T_u(\phi)=\int_\Omega u\phi\, dx.$$ If you now have a sequence of functions $u_n$ in $L^p(\Omega)$ converging to some function $u$ in $L^p(\Omega)$, and you consider the corresponding distributions $T_{u_n}$, then as Dap said, using Hölder's inequality you get $$|T_{u_n}(\phi)-T_{u}(\phi)|=\left|\int_\Omega (u_n-u)\phi\, dx\right|\le \Vert u_n-u\Vert_{L^p}\Vert \phi\Vert_{L^{q}}\to 0$$ as $n\to\infty$. Thus $T_{u_n}$ converges to $T_{u}$ in the sense of ditributions. Since each $\phi$ has compact support, it would be enough to assume that $u_n$ converges to $u$ in $L^p(K)$ for every compact set $K\subset\Omega$.

Gio67
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  • Or integrate $u_n \to u\in L^1_{loc}$ to obtain a locally uniformly convergent sequence of continuous functions, and say that convergence in $C^0$ implies convergence of all the distributional derivatives. – reuns Oct 01 '17 at 02:26
  • @Gio67 : I got what you said, thanks. – Mathslover shah Oct 02 '17 at 04:02
  • @reuns : A bit confused about your idea ??? – Mathslover shah Oct 02 '17 at 04:03
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    great. If I answered your question, then you should accept my answer. Thanks! – Gio67 Oct 02 '17 at 11:16
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    @Mathslovershah If $u_n \in L^p, p \ge 1$ then $v_n(x) = \int_0^x u_n(y)dy $ is continuous and $v_n \to v$ locally uniformly, which implies it converges in the sense of distributions and the same for all the distributional derivatives $v_n' = u_n \to u = v'$ – reuns Oct 02 '17 at 15:02