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Yesterday assistant filed during exercises work, but not one of us was able to resolve, and then he proved himself and noticed that he had the solution when attempting to problems. Therefore, please know that there are many people who can solve this example. so many to choose you please, but in detail, please. The example is as follows:

Let $X$ and $Y$ are normed spaces. By $L_c(X,Y)$ we denote the set of all compact operators from $X$ to $Y$. Further we put $L_c(X)=L_c(X,X)$.

IF $X$ is a normed spaces, then $L_c(X)$ two sided ideal in normed algebra $L(X)=L(X,X)$.

Previously thank you from the heart

user145717
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This question is not too hard provided one uses the right definiton of compact operator. The one which works best here (for me at least) is that an operator $C:X \to X$ is compact if and only if it takes sequences in the unit ball in $X$ into sequences having Cauchy convergent subsequences. This is one of the several equivalent formulations given in the wikipdia page for compact operators, http://en.m.wikipedia.org/wiki/Compact_operator. To see it applies in this case, we employ it to show that, if $C$ is compact, then both $AC$ and $CA$ are compact for any bounded $A \in L(X)$, and this will establish that $L_c(X)$ is in fact a two-sided ideal in $L(X)$, the algebra of bounded operators on $X$.

Before proceding with the details of the proof, however, we note that the definition may be expanded slightly to affirm that compactness is equivalent to the hypothesis that $A$ maps sequences in any ball to sequences with Cauchy subsequences. This is clearly a logical consequence of the wikipedia definition, since a sequence $x_j \in \bar B(0, R)$, the closed ball of radius $R$ about $0$, may be scaled by $R^{-1}$ to become a sequence $R^{-1} x_j \in \bar B(0, 1)$; then $A(R^{-1} x_j)$ has a Cauchy subsequence, and hence, scaling such a subsequence by $R$, we see that $R(A(R^{-1} x_j)) = Ax_j$ has a Cauchy subsequence as well. So the two variants of the definition are equivalent.

We use these observations to produce the desired result. If $A \in L(X)$ and $C \in L_c(X)$, and $x_j \in \bar B(0, 1)$, then $Cx_j$ has a Cauchy subsequence and so, by continuity (since $A$ is bounded), does $ACx_j$ (simply take the image of a Cauchy subsequence of $Cx_j$ under $A$). This shows $AC$ is compact, so $L_c(X)$ is a left ideal. To see that it is also a right ideal, consider the sequence $Ax_j$, again with $x_j \in \bar B(0, 1)$. Since for every $j$ $\Vert Ax_j \Vert \le \Vert A \Vert \Vert x_j \Vert \le \Vert A \Vert$, we see that the sequence $Ax_j \in \bar B(0, \Vert A \Vert)$, whence according the "enhanced definition" of compact $C$ we have just developed, $CAx_j$ contains a Cauchy subsequence, so that $CA$ is compact. Thus $L_c(X)$ is a right ideal in $L(X)$ as well as a left ideal; it is thus two-sided, as was to be shown. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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