# Bounded operator to a space with two different norms

$$X$$ 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.

## 1 Answer

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.