The question is just like the title. (For $E$ measurable and $1\le p<∞$, define $L^p(E)$ to be the collection of measurable functions $f$ for which $|f|^p$ is integrable over $E$; thus $L^1(E)$ is the collection of integrable functions.)
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Let $X = [1, \infty]$, $f_n(x) = \dfrac{1}{nx} + 1$ and $f(x) = 1$. We have: $$ \|f_n - f\|_2 = \left\{\int_1^\infty \left|\dfrac{1}{nx}\right|^2\,dx\right\}^{1/2} = \dfrac{1}{n} $$ Therefore, $f_n \to f$ in $L^2([1, \infty])$. On the other hand: \begin{align*} \|f_n^2 - f^2\|_1 &= \int_1^\infty \left|\left(\dfrac{1}{nx} + 1\right)^2 - 1\right| \,dx \\ &= \int_1^\infty \left|\left(\dfrac{1}{nx}\right)^2 + \dfrac{2}{nx}\right| \,dx \\ &\ge \dfrac{2}{n} \int_1^\infty \dfrac{1}{x} \,dx \end{align*} Which diverges no matter what $n$ is. |
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In a question of this type, always check $f(x)=1/x$ in the sets $E_1=(0,1)$ and $E_2=(1,\infty)$. Notice that $f(x)$ is not integrable on either set in $L^1$, but it is integrable on one of them in $L^p$. |
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If $E \subset \mathbb R^n$ is bounded, this is true if I'm not mistaken. Consider the superposition operator corresponding to $\operatorname{id}^p$, i.e. the operator $$T \colon L^p(E) \to L^1(E)$$ with $$(Tg)(x) = g(x)^p$$ for every $g \in L^p(E)$. A well-known result for superposition operators (see e.g. Theorem 1.2.1 in Progress In Nonlinear Differential Equations and Their Applications or Remark 2.5 in A Primer of Nonlinear Analysis) says that, because of its limited growth and pointwise continuity, $T$ is continuous. In other words, $f_n^p = Tf_n \to Tf = f^p$ in $L^1$. |
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