I had this exercise on a functional analysis exam but I was unable to solve point iii). I solved points i), ii), and iv). I solved i) with the open map theorem, ii) using i) and iv) as an instance of iii). I wanted to use ii) to solve iii) but I can't do that because I don't know if $Y$ is a Banach space with the norm $\|\cdot\|_1$. I also proved that if $\hat Y \subset Y$ is closed with respect to the norm $\|\cdot\|_1$ then it is closed with respect to the norm $\|\cdot\|_2$ and so is a Banach space with $\|\cdot\|_1$ and $\|\cdot\|_2$. In this last case, I can apply ii) to prove that $T|_\hat Y$ is $(\|\cdot\|_2, \|\cdot\|_2)$-bounded.
Here is the exercise:
i) Let $X$ be a vector space and let $\|\cdot\|_1$, $\|\cdot\|_2$ be norms on $X$. Assume that $X$ is complete with respect to both norms and that there exists a constant $C > 0$ such that $\|x\|_1 \leq C\|x\|_2$ for all $x \in X$. Show that there exists $c > 0$ such that $\|x\|_2 \leq c\|x\|_1$ for all $x \in X$, and therefore the two norms are equivalent.
[Hint: Consider the identity operator $I : (X, \|\cdot\|_2) \rightarrow (X, \|\cdot\|_1)$]
ii) Let $(X, \|\cdot\|)$ and $(Y, \|\cdot\|_1)$ be Banach spaces and let $\|\cdot\|_2$ be a second norm on $Y$ with respect to which $Y$ is complete and such that $\|\cdot\|_1 \leq C\|\cdot\|_2$ for some $C > 0$. Finally, let $T : X \rightarrow Y$ be a linear operator. Show that if $T$ is $(\|\cdot\|, \|\cdot\|_1)$-bounded, that is, $\|T x\|_1 \leq M \|x\|$ for some $M > 0$ and all $x \in X$, then $T$ is also $(\|\cdot\|, \|\cdot\|_2)$-bounded.
iii) Let $(X, \|\cdot\|_1)$ be a Banach space and let $Y$ be a vector subspace of $X$ which is itself a Banach space with respect to a second norm $\|\cdot\|_2$ on $Y$ for which $\|y\|_1 \leq C\|y\|_2$ for all $y \in Y$. Suppose that $T : X \rightarrow X$ is linear, $(\|\cdot\|_1, \|\cdot\|_1)$-bounded and that $T(Y) \subset Y$. Show that $T|_Y : Y \rightarrow Y$ is also $(\|\cdot\|_2, \|\cdot\|_2)$-bounded.
iv) Let $T : L^2([0, 1]) \rightarrow L^2([0, 1])$ be a linear bounded operator with respect to the usual $L^2$-norm. Assume that $Tf \in C([0, 1])$ for every $f \in C([0, 1])$. Show that $T : C([0, 1]) \rightarrow C([0, 1])$ is bounded with respect to the uniform norm $\|\cdot\|_{\infty}$.
Can you give me a hint to solve the point iii)?
Since you asked only for a hint: Apply the Closed Graph Theorem to the operator $T: (Y, \|\cdot\|_2) \to (Y, \|\cdot\|_2)$ by using the fact that the assumptions tell you that convergent sequences in $\|\cdot\|_2$ are convergent in $\|\cdot\|_1$ to the same limit to verify the hypotheses of that result.