# $X$ be a real normed linear space ; if $\mathcal L(X,X)$ is complete then is $X$ also complete?

Let $X$ be a real normed linear space and $\mathcal L(X,X)$ denote the set of all bounded linear operators on $X$ , we know that if $X$ is complete then so is $\mathcal L(X,X)$ ; is the converse true i.e. if $\mathcal L(X,X)$ is complete then is $X$ also complete ?

If $X = \{0\}$, then $X$ is complete, as is $\mathcal{L}(X,X)$. If $X \neq \{0\}$, choose $\lambda \in X'$ with $\lVert\lambda\rVert_{X'} = 1$.

Then $\iota\colon x \mapsto \lambda \otimes x$, where $(\lambda \otimes x)(y) = \lambda(y)\cdot x$ is an isometric embedding of $X$ into $\mathcal{L}(X,X)$, and $\iota(X)$ is a closed subspace.

The linearity of $\iota$ is clear, and we have

\begin{align} \lVert \iota(x)\rVert_{\mathcal{L}(X,X)} &= \sup \{ \lVert \iota(x)(y)\rVert_X : \lVert y\rVert_X \leqslant 1\}\\ &= \sup \{ \lVert \lambda(y)\cdot x\rVert_X : \lVert y\rVert_X \leqslant 1\}\\ &= \sup \{ \lvert \lambda(y)\rvert\cdot \lVert x\rVert_X : \lVert y\rVert_X \leqslant 1\}\\ &= \bigl(\sup \{ \lvert \lambda(y)\rvert : \lVert y\rVert_X \leqslant 1\}\bigr)\cdot\lVert x\rVert_X\\ &= \lVert\lambda\rVert_{X'}\cdot \lVert x\rVert_X\\ &= \lVert x\rVert_X, \end{align}

so $\iota$ is an isometry. Next we note that $T \in \iota(X) \iff \ker \lambda \subseteq \ker T$. If $T = \iota(x)$, then $\ker\lambda \subseteq \ker T$ is clear.

Conversely, suppose $\ker \lambda \subseteq \ker T$. Choose $x_1 \in X$ with $\lambda(x_1) = 1$. Since $\lambda \neq 0$, such an $x_1$ exists. Then we have $T = \iota(T(x_1))$: We can write every $x\in X$ as $x = \lambda(x)\cdot x_1 + (x - \lambda(x)\cdot x_1)$, where $x - \lambda(x) \cdot x_1 \in \ker \lambda$. Then

$$T(x) = T\bigl(\lambda(x)x_1 + (x - \lambda(x)x_1)\bigr) = \lambda(x)T(x_1) + T(\underbrace{x-\lambda(x)x_1}_{\in \ker \lambda \subseteq \ker T}) = \lambda(x)T(x_1) = \iota(T(x_1))(x),$$

so $T = \iota(T(x_1))$.

Finally, we deduce that $\iota(X)$ is closed in $\mathcal{L}(X,X)$ from this characterisation: If $T \notin \iota(X)$, then $\ker \lambda \not\subseteq \ker T$, so there is an $y \in \ker \lambda \setminus \ker T$. For $\lVert T - S\rVert_{\mathcal{L}(X,X)} < \frac{\lVert T(y)\rVert_X}{\lVert y\rVert_X}$ we then have

$$\lVert S(y)\rVert_X = \lVert T(y) - (T-S)(y)\rVert_X \geqslant \lVert T(y)\rVert_X - \lVert T-S\rVert_{\mathcal{L}(X,X)}\cdot \lVert y\rVert_X > \lVert T(y)\rVert_X - \lVert T(y)\rVert_X = 0,$$

so $S(y) \neq 0$ and therefore $\ker \lambda \not\subseteq \ker S$, whence $S\notin \iota(X)$, showing that $\mathcal{L}(X,X)\setminus \iota(X)$ is open.

Thus, for normed spaces, we have $X\text{ complete} \iff \mathcal{L}(X,X) \text{ complete}$.

• @Daniel Fischer♦ : what do you mean by $X'$ ? – user228168 Feb 15 '16 at 16:10
• Per the question, it's the space of bounded linear operators on $X$, i.e., since $X$ is normed, the space of continuous linear maps $X\to X$. Although it's not explicitly stated, I assumed that it's endowed with the operator norm, @user1952009. – Daniel Fischer Feb 15 '16 at 16:10
• @SaunDev The topological dual of $X$. Maybe you're used to the notation $X^{\ast}$? – Daniel Fischer Feb 15 '16 at 16:10
• @Daniel Fischer : ok, but I'm missing something..! as $X$ is dense in its completion, I can't see how the set of bounded operators $X \to \mathbb{K}$ or the set of bounded operators $X \to X$ can be different for $X$ complete or not. – reuns Feb 15 '16 at 16:16
• @Svetoslav Let $K = \ker \lambda$. Then $T \in \iota(X) \iff K \subseteq \ker T$. Then it's easy to show that the complement of $\iota(X)$ is open. (The completeness of $\mathcal{L}(X,X)$ doesn't play a role for that, $\iota(X)$ is always closed.) – Daniel Fischer Feb 15 '16 at 16:39