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Given two Banach spaces $(X,\Vert \cdot\Vert_X)$ and $(Y,\Vert \cdot\Vert_Y)$ such that $X\subset Y$ with continuous embedding (i.e. there exits a constant $c>0$ such that $\Vert x\Vert_Y \leq c\Vert x\Vert_X$), we consider an operator $T\in \mathcal{L}(L^2([0,1],X);L^2([0,1],Y))$ which verifies $(Tf)(t) \in X$ for almost every $t\in [0,1]$. My question is : Can i say that $T$ is bounded on $L^2([0,1],X)$, i.e. $T\in \mathcal{L}(L^2([0,1],X);L^2([0,1],X))$? Thanks in advance.

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If both norms are equivalent, then the answer is yes. Otherwise: no. Assume that both norms are not equivalent.

Then for each $n$ there is $x_n\in X$ with $\|x_n\|_X > n \|x_n\|_Y$. Wlog we can assume $\|x_n\|_Y=1$. Define $$ u := \sum_{n=1}^\infty \chi_{(\frac1{n+1},\frac1n)}x_n . $$ Then $$ \|u\|_{L^2(I,Y)}^2 = \sum_{n=1}^\infty \frac1{(n+1)^2} <+\infty $$ and $$ \|u\|_{L^2(I,X)}^2 \ge \sum_{n=1}^\infty \frac{n^2}{(n+1)^2} =+\infty. $$ Hence $u\in L^2(I,Y)\setminus L^2(I,X)$. Take $F\in L^2(I,X)^*\setminus\{0\}$. Then $$ Tv:= F(v)\cdot u $$ is a counterexample to the claim.

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