# If $T_1, T_2$ are linear compact operators, then $T_1+T_2$ is also linear and compact.

Let $$X,Y$$ be Banach space.

Prove that if $$T_1 : X\to Y, T_2 : X\to Y$$ are both linear and compact operators, then $$T_1+T_2$$ is also linear and compact.

The definition of compact operator is here.

Compact Operator

$$T:X\to Y$$ is compact iff for every bounded sequence $$\{x_n\}_{n=1}^\infty\subset X$$, there exists a subsequence of $$\{Tx_n\}_{n=1}^\infty$$ converging to an element in $$Y.$$

Let me prove.

The linearity of $$T_1+T_2$$ is almost obvious from the linearity of $$T_1,T_2$$ and the definition of sum of operator, so I should prove the compactness of $$T_1+T_2.$$

Let $$\{x_n\}_{n=1}^\infty\subset X$$ be bounded.

Since $$T_1$$ is compact, there exist

$$\{T_1 y_n\}_{n=1}^\infty\$$ : subsequence of $$\{ T_1 x_n \}_{n=1}^\infty\ \$$ and $$\lambda \in Y$$

such that $$T_1 y_n\to \lambda$$ as $$n\to \infty.$$

Now, $$\{y_n \}$$ is a subsequence of $$\{x_n\}$$ and since $$\{x_n\}$$ is bounded, $$\{y_n\}$$ is also bounded.

From the compactness of $$T_2$$, there exist

$$\{T_2 z_n\}_{n=1}^\infty\$$ : subsequence of $$\{ T_2 y_n \}_{n=1}^\infty\ \$$ and $$\mu \in Y$$

such that $$T_2 z_n \to \mu$$ as $$n\to \infty$$.

Then, $$(T_1+T_2)z_n=T_1 z_n +T_2z_n \to \lambda +\mu\in Y,$$ where $$T_1 z_n\to \lambda$$ is from the fact that $$\{T_1 z_n\}$$ is a subsequence of $$\{T_1 y_n\}$$ and $$T_1 y_n\to \lambda,$$ and $$\lambda+\mu\in Y$$ is from the fact that $$Y$$ is vector space.

This shows that $$\{ (T_1+T_2)z_n\}_{n=1}^\infty$$ is a subsequence of $$\{ (T_1+T_2)x_n\}_{n=1}^\infty$$ converging to an element of $$Y,$$ thus $$T_1+T_2$$ is compact.

Does this proof make sense ?

Please point out if there are any mistakes and/or errors.

• Surely it would be easier to prove from the definition? If $A,B$ are relatively compact then $A+B$ is relatively compact. Jul 19, 2022 at 16:31
• Looks good to me. However, I object to using $\lambda,\mu$ to represent vectors. You don't want to veer from accepted standards in the subject or you will confuse people. Jul 25, 2022 at 14:48

$$T\in \mathcal{L}(X, Y)$$ is compact iff $$T(\overline{B(0, 1) }) \subset Y$$ is compact.
$$(T_1+T_2)(\overline{B(0, 1) })=T_1(\overline{B(0, 1) })+T_2(\overline{B(0, 1) })$$
Now it follows from Compactnesss of $$T_1$$ and $$T_2$$ that $$T_1+T_2$$ is compact.