Let $l^2$ be the Hilbert space of infinite sequences $(x_1, x_2, ...)$ such that $\|x\|^2 = \sum_{k=0}^{\infty} |x_k|^2 < \infty$. Define a function $M: l^2 \to l^2$ so that $$Mx := (m_1x_1, m_2x_2, m_3x_3, ...)$$
- What restrictions on the $m_k$ do we need for $M$ to be a linear transformation?
- What restrictions do we need for $M$ to be one-to-one?
- What restrictions do we need for $M$ to be onto?
- Suppose $M$ is bijective. How can we find its inverse $M^{-1}$ and what's the bound on the norm of this inverse?
For $M$ to be a linear transformation, we only need $M(cx + ky) = cM(x) + kM(y)$. But since $M$ just multiplies each entry of an element of $l^2$ by some number $m_k$, it doesn't look like there has to be any restriction since $$(m_1(cx_1 +ky_1), .....) \to ((m_1cx_1 + m_1ky_1),...) \to ((m_1cx_1),...)+((m_1ky_1),...) \to c(m_1x_1, ...) + k(m_1y_1, ...)$$
Next, for $M$ to be one-to-one, my guess would be that all the $m_k$ cannot be zero, so that $N(M) = \{0\}$
I don't know why, but I have no clue about how to make sure $M$ is onto
Lastly, if $M$ is bijective, I believe we could define $M^{-1}$ as $$M^{-1}x := (\frac{1}{m_1}x_1, \frac{1}{m_2}x_2, \frac{1}{m_3}x_3 ...)$$
Then to find a bound of the norm, I think I should use some aspect of the fact that $l^2$ is a Hilbert space, but I'm not sure how.
Thank you for your input!
Hints:
(a) An important and easy to forget condition is $Mx\in l^2$ (for all $x\in l^2$).
(b) Your ideas about when $M$ is one-to-one and the formula for $M^{-1}$ are correct. Think again in (a)
(c) The condition required for (b) is also necessary for $M$ being onto, because if some $m_i = 0$ then the $i$-th coordinate of $Mx$ always will be zero and the image will be strictly smaller than $l^2$.
(d) About the norm:
$$\|Mx\|^2 = \sum _{k=1}^\infty|m_kx_k|^2\le\cdots$$ (some obvious bound will be...?)