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Let $X,Y$ be normed vector spaces over $\mathbb{C}$, and $L(X,Y)$ the space of all bounded linear maps from $X$ to $Y$. Its known that $L(X,Y)$ is a normed(operator norm) vector space.

Theorem:

The completeness of $Y$ implies the completeness of $L(X,Y)$ in the norm metric.

Proof: Let $(f_j)_{j=0}^\infty$ be a Cauchy sequence in $L(X,Y)$. If $\mathbf{x} \in X$, then for indices $j_0,j_1$, $\lVert f_{j_0}(\mathbf{x}) - f_{j_1}(\mathbf{x}) \rVert \leq \lVert f_{j_0} - f_{j_1} \rVert \lVert \mathbf{x} \rVert$. Hence $(f_j(\mathbf{x}))_{j=0}^\infty$ is Cauchy.

Is the following proof justified?

Define $f:X \to Y$ by $f(\mathbf{x}) = \lim_{j \to \infty} f_j(\mathbf{x})$. To show the linearity of $f$, $$\begin{align*} f(\mathbf{x} + \lambda\mathbf{y}) &= \lim_{j \to \infty} f_j(\mathbf{x} + \lambda\mathbf{y}) \\ &= \lim_{j \to \infty} f_j(\mathbf{x}) + \lambda \lim_{j \to \infty} f_j(\mathbf{y}) \\ &= f(\mathbf{x}) + \lambda f(\mathbf{y}) \end{align*}$$ To show the boundedness of $f$, since every $f_j$ is bounded, $\forall j \in \mathbb{W}, \exists c_j > 0\mid \forall \mathbf{x} \in X, \lVert f_j(\mathbf{x}) \rVert \leq c_j \lVert \mathbf{x} \rVert$. We can let this $c_j$ be $\lVert f_j \rVert$(this is a known fact), and $$\begin{align*} \lVert f(\mathbf{x}) \rVert &= \lim_{j \to \infty} \lVert f_j(\mathbf{x}) \rVert \\ &\leq \lim_{j \to \infty} \lVert f_j \rVert \lVert \mathbf{x} \rVert \\ &= \lVert \mathbf{x} \rVert \lim_{j \to \infty} \lVert f_j \rVert \end{align*}$$

How do I show the existence of $\lim_{j \to \infty} \lVert f_j \rVert$?

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  • $\begingroup$ the limit $T$ of a converging sequence of (uniformely) bounded operators is bounded, and it is an operator from $X \to \bar{Y}$ so if $Y$ is complete ($\bar{Y} = Y$) you get $T \in L(X,Y)$, I don't see the point. $\endgroup$ – reuns Feb 6 '16 at 6:36
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Note that for any normed linear space $X$, $f, g, \in X$,

$$\left| \|f\| - \|g\| \right| \le \|f-g\|$$

by triangle inequality. So in your situation $\|f_j\|$ is a Cauchy sequence in $\mathbb R$.

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  • $\begingroup$ How do I show that $\lVert f \rVert = \lim_{j \to \infty} \lVert f_j \rVert$? $\endgroup$ – Henricus V. Feb 6 '16 at 6:20
  • $\begingroup$ @Henry W : the assumption is that the sequence is converging : $||f - f_j|| \to 0$ $\endgroup$ – reuns Feb 6 '16 at 6:34

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