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I have already read some similar questions on stack exchange but couldn't find an answer to my question. So I hope it's okay I ask this:

Prove: $f: \ell^1 \rightarrow (c_0)'$, $f_y(x)= \sum \limits_{i=1}^\infty x_i*y_i\ $ is an isometric isomorphism. Does is also work if you take $\ell^\infty$ instead of $\ell^1$?

Where $(c_0)'$ is the dual space of $c_0$, the space of all zero - sequences in $\ell^\infty$ and y is a sequence in $c_0$.

I managed to show that $f$ is bounded and linear, but don't know how to prove the injectivity and surjectivity of f. I found a question where it was proved that $\ell^1 = (c_0)'$, but that doesn't show that f must be bijective, does it?

Plus, how can I show that $\lVert f_y \rVert = \Vert y \rVert$?

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You have boundedness and linearity, so let's check that it's a surjective isometry. Then injectivity will follow, for if $f$ is linear, then $\|fx\|=0$ if and only if $\|x\| = 0$.

Let's do surjectivity first. Given a bounded linear functional $g$ on $c_0$, can we find a sequence in $\ell^1$ that represents it? Sure. Define $y_n = (sgn(g(e_1)), sgn(g(e_2)), \ldots sgn(g(e_n)), 0, 0, \ldots)$ where the $e_i$ are the Schauder basis for $\ell^1$. Then $y \in c_0$ with $\|y\| =1$, so $\|g\| \ge \displaystyle \sum_{i=1}^n |g(e_i)|$ for all $n$. Letting $n \to \infty$, we see the sequence $y$ defined by $y_i =g(e_i)$ is in $\ell^1$. And this is exactly what we want. For we know we can write $x= (x_1, x_2, \ldots) = \displaystyle \sum_{i=1}^\infty x_ie_i$ for $x \in c_0$. Then $g(x) = \sum_{i=1}^\infty x_i g(e_i)$. So surjectivity is done.

For the isometry, we have $\|g\| \ge \|y\|$ by the same argument above, noting that $\|y_n\| \le \|g\|$ for all $n$. But for $x \in c_0$ with $\|x\|=1$ we have that $\|g\| = \sup_{\|x\|=1} \left|\sum_{i=1}^\infty x_ig(e_i)\right|= \sup_{\|x\|=1} \left|\sum_{i=1}^\infty x_iy_i\right| \le \sup_{\|x\|=1}\sum_{i=1}^\infty |x_iy_i| \le \sup_{\|x\|=1} \|x\|\sum_{i=1}^\infty |y_i| = \|y\|$, so $\|g\| = \|y\|$.

Thus $f$ is an isometry. Since you've already shown $f$ is a bounded linear operator, we can conclude $f: \ell^1 \to (c_0)^*$ is an isometric isomorphism.

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  • $\begingroup$ Thank you a lot for helping me! It's a really nice proof! $\endgroup$ – Infinite_28 Jun 4 '17 at 12:37
  • $\begingroup$ Could you help me with the second question, i.e. why does the proof not work when taking $(\ell^\infty)'$ instead of $(c_0)'$? I've read that $ \ell^1 \subsetneq (\ell^\infty)'$, so I know there cannot be a bijective linear mapping. But where in the proof can I see it doesn't work? $\endgroup$ – Infinite_28 Jun 4 '17 at 13:03
  • $\begingroup$ I'd say you cannot take $\ell^\infty$ instead of $c_0$, because $g(e_i) \in \ell^\infty$ does not mean that $g(e_i)$ has to be in $\ell^1$. That's the point, isn't it? $\endgroup$ – Infinite_28 Jun 5 '17 at 17:27

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