Let $T_1: X \rightarrow Y$ be a closed linear operator and $T_2: X \rightarrow Y$ a bounded linear operator and $X$ and $Y$ normed spaces over the same field.

Is the sum of such operators also closed?

Here is my reasoning so far:

If we take a convergent sequence in the graph $G(T_1 + T_2)$, assuming such sequence exists, say $z_n = (x_n, (T_1+T_2)x_n)$ then it has to converge in their own pair of spaces, meaning, $x_n \rightarrow x \in X$ and $(T_1 +T_2)x_n \rightarrow (T_1 +T_2)x \in R(T_1+T_2) \subset Y$

Now, it seems that boundedness of $T_2$ is what guarantees that the the graph $G$ above will be closed. (Obviously along with linearity).

In this context, boundedness implies continuity and if there exists such $x_n$ then it has to converge to a point in the range of $(T_1+T_2)$.

I am having trouble understanding how I could use this fact though, namely $\|T_2x\| \leq k\|x\|$.

I would really appreciate hints on how to do so. Thanks in advance!


Take $T_1$ closed, $T_2$ bounded and $x_n\to x \in X$ so that $(T_1+T_2)x_n \to y\in Y.$ Then $T_1x_n\to y-T_2x$. (That step is where you use $T_2$ being continuous). $T_1$ is closed so $T_1x=y-T_2x$ and rearranged you have your result

  • $\begingroup$ Appreciate it, @enthdegree! $\endgroup$ – upStoneLock May 21 at 22:12

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