Given that $X$ and $Y$ are normed linear spaces and that $T:X\to Y$ is a linear map, such that $T(\alpha x_1 + \beta x_2) = \alpha T(x_1)+ \beta T(x_2)$ for all vectors $x_1,x_2\in X$ and scalars $\alpha , \beta$. Suppose $T$ maps $X$ onto $Y$ and is isometric, $\|Tx\|=\|x\|$ for all $x\in X$.

1) Show that if $X$ is a Hilbert space then so is $Y$ if we define:

$$\langle y_1,y_2\rangle_Y=\langle x_1,x_2\rangle_X$$

Where $x_1,x_2$ are the unique points in $X$ satisfying $Tx_1=y_1$ and $Tx_2=y_2$

What I have done so far

I have previously proved that if $X$ is a Banach space then so is $Y$.

So I am thinking that since $X$ is a Hilbert space,

$$||y_1-y_2|| = ||T(x_1)-T(x_2)|| = ||T(x_1-x_2)||=||x_1-x_2|| \implies$$

$$||y_1-y_2||^2 = ||y_1||^2 - 2Re\langle y_1,y_2 \rangle + ||y_2||^2$$

$$||x_1-x_2||^2 = ||x_1||^2 - 2Re\langle x_1,x_2 \rangle + ||x_2||^2$$


$$||y_1+y_2|| = ||T(x_1)+T(x_2)|| = ||T(x_1+x_2)||=||x_1+x_2|| \implies$$

$$||y_1+y_2||^2 = ||y_1||^2 + 2Re\langle y_1,y_2 \rangle + ||y_2||^2$$

$$||x_1+x_2||^2 = ||x_1||^2 + 2Re\langle x_1,x_2 \rangle + ||x_2||^2$$



Therefore the parallelogram equality holds for $Y$, and therefore there is an inner product that gives the norm, and so $||y_1 - y_2||=\sqrt{\langle y_1-y_2,y_1-y_2 \rangle}$ and therefore $Y$ is a Hilbert space.

  • $\begingroup$ Good job! Carry on. $\endgroup$ – Mhenni Benghorbal Sep 12 '14 at 20:07

You should go this way

$$ ||y_1-y_2|| = ||T(x_1)-T(x_2)|| = ||T(x_1-x_2)||=||x_1-x_2||$$

since you have been given

$$ ||T(x)||=||x|| $$

and hence the parallelogram law will hold.

  • $\begingroup$ So using the above, I can do the same for $\|y_1+y_2\|$ then, and from there get the parallelogram equality, correct? Also, I know that if the parallelogram equality holds that it means there is an inner product that gives the norm, so does that conclude that $Y$ is a Hilbert space? $\endgroup$ – user3784030 Sep 12 '14 at 3:43
  • $\begingroup$ @user3784030: The norm in Hilbert space is defined by the inner product. By the way you can do the same for $||y_1+y_2||$. $\endgroup$ – Mhenni Benghorbal Sep 12 '14 at 3:53
  • $\begingroup$ So how does this use the fact that $\langle y_1,y_2\rangle_Y=\langle x_1,x_2\rangle_X$ needs to be defined? $\endgroup$ – user3784030 Sep 12 '14 at 4:14
  • $\begingroup$ @user3784030: In $(1)$ It is already defined that $\langle y_1,y_2\rangle_Y=\langle x_1,x_2\rangle_X$. Just check the parallelogram law. $\endgroup$ – Mhenni Benghorbal Sep 12 '14 at 4:20
  • $\begingroup$ Could you let me know how the update looks, @mhenni-benghorbal $\endgroup$ – user3784030 Sep 12 '14 at 5:00

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