$X^*$ is a Banach space with the operator norm iff $X$ is a Banach space

Question

is this statement true or false?

"$X^*$ is a Banach space with the operator norm iff $X$ is a Banach space"

$X^*$ is the dual space of $X$

can someone explain thanks.

Answer 1

The statement is false in general.

The dual space $X^*$ of any normed space $X$ is complete.

Proof:

Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $X$. For any $x \in X$ we have:

$$\|f_m(x) - f_n(x)\| = \|(f_m - f_n)(x)\| \le \|f_m - f_n\| \|x\| \xrightarrow{m, n \to \infty} 0$$

Hence, $(f_n(x))_{n=1}^\infty$ is a Cauchy sequence for all $x \in X$. Since the field $\mathbb{F}$ is complete, all the sequences converge in $\mathbb{F}$.

Define $f : X \to \mathbb{F}$ as $$f(x) = \lim_{n\to\infty} f_n(x)$$ for all $x \in X$.

$f$ is linear:

$$f(\alpha x + \beta y) = \lim_{n\to\infty} f_n(\alpha x + \beta y) = \lim_{n\to\infty} \left(\alpha f_n(x) + \beta f_n(y)\right) = \alpha \lim_{n\to\infty} f_n(x) + \beta \lim_{n\to\infty} f_n(y) = \alpha f(x) + \beta f(y)$$

Let's prove that $f$ is bounded and that $f_n \xrightarrow{n\to\infty} f$ in the operator norm. Take $\varepsilon > 0$ and fix $n_0 \in \mathbb{N}$ such that for all $m, n \ge n_0$ we have:

$$\|f_m - f_n\| < \frac\varepsilon2$$

So for any $x \in X$ we have:

$$\|f_m(x) - f_n(x)\| = \|(f_m - f_n)(x)\| \le \|f_m - f_n\|\|x\| < \frac\varepsilon2 \|x\| $$

Letting $m\to\infty$ we obtain that for all $n \ge n_0$ holds:

$$\|(f - f_n)(x)\| = \|f(x) - f_n(x)\| \le \frac\varepsilon2 \|x\|$$

Hence the linear map $f - f_n$ is bounded for all $n \in \mathbb{N}$. This implies that $f$ is also bounded since $f = (f - f_n) + f_n$ and $X^*$ is a vector space.

Finally, taking the supremum over all $\|x\| \le 1$ we get that for all $n \ge n_0$ holds:

$$\|f - f_n\| \le \frac\varepsilon2 < \varepsilon$$

so $\|f - f_n\| \xrightarrow{n\to\infty} 0$, which shows $f_n \xrightarrow{n\to\infty} f$.

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Now we see that the converse cannot hold. Take any non-complete normed space, for example $c_{00}$. According to the previous discussion, $(c_{00})^*$ is complete, and $c_{00}$ is not.