$X$ is a Banach space and $Y$ is a normed linear space. $(Y,\lVert\cdot\rVert_1)$ is not complete and $(Y,\lVert\cdot\rVert_2)$ is complete, while $\lVert\cdot\rVert_2\ge\lVert\cdot\rVert_1$. Let $T: X\to( Y,\lVert\cdot\rVert_1)$ be a bounded linear operator. Prove that $T: X\to( Y,\lVert\cdot\rVert_2)$ is also a bounded linear operator.
For example, $Y$ can be $C([0,1])$. $\lVert\cdot\rVert_1$ is $L_1$ norm and $\lVert\cdot\rVert_2$ is sup norm.
I've found a hint from my textbook telling me using closed graph theorem, so I try to give a proof now. By closed graph theorem, all I need is to prove T is a closed operator.
First, there is a lemma which helps, and the proof can be easily found on Wiki.
Lemma:$T:X\to Y$, for any sequence $x_n\to x_0$ and $Tx_n=y_n\to y_0$, if $x_0\in\mathfrak D(T)$ and $y_0=Tx_0$, then $T$ is a closed operator.
Let $x_n$ and $y_n$ be the same as in the lemma. $\lVert Tx_0-y_0\rVert_1 \leq\lVert Tx_0-Tx_n\rVert_1+\lVert Tx_n-y_0\rVert_1$. Then, $\lVert Tx_0-y_0\rVert_1 \leq \lim_{n\to \infty} C\lVert x_0-x_n\rVert_X+\lVert Tx_n-y_0\rVert_2=0$. Hence $Tx_0=y_0$ and T is a closed operator, which completes the proof.
