Linearity of Sums and Scalar Multiples of Square Lebesgue Integrable Functions
Recall from the Square Lebesgue Integrable Functions page that a function $f$ is said to be square Lebesgue integrable on an interval $I$ if $f$ is a Measurable function and $f^2$ is Lebesgue integrable on $I$. Furthermore, the set of all square Lebesgue integrable functions on $I$ is denoted $L^2 (I)$.
On the Products of Square Lebesgue Integrable Functions are Lebesgue Integrable page we looked at a nice theorem which said that if $f$ and $g$ are square Lebesgue integrable functions on an interval $I$ then $fg$ is Lebesgue integrable on $I$.
We will now prove that the sum of two square Lebesgue integrable functions on $I$ is square Lebesgue integrable on $I$ and that any scalar multiple of a square Lebesgue integrable function on $I$ is square Lebesgue integrable on $I$.
| Theorem 1 (Additivity): Let $f$ and $g$ be square Lebesgue integrable functions on $I$. Then $f + g$ is square Lebesgue integrable on $I$. |
- Proof: If $f$ and $g$ are both square Lebesgue integrable functions on $I$ then $f, g \in M(I)$ and $f^2, g^2 \in L(I)$.
- Since $f, g \in M(I)$ there exists sequences of step functions $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ that converge to $f$ and $g$ (respectively) almost everywhere on $I$. Define a new sequence of functions as $(f_n(x) + g_n(x))_{n=1}^{\infty}$. Then this sequence is a sequence of step functions (since the sum of two step functions is a step function) that converges to $f + g$ almost everywhere on $I$. Therefore $(f + g) \in M(I)$.
- Now we also have that $f^2, g^2 \in L(I)$ and we want to show that $(f + g)^2 \in L(I)$. Notice that:
- Since $f, g \in L^2(I)$ we have by the theorem referenced above that $fg \in L(I)$. So by Linearity of Lebesgue Integrals page we see that $(f^2 + 2fg + g^2) \in L(I)$ which shows that $(f + g)^2 \in L(I)$.
- Therefore $(f + g) \in M(I)$ and $(f + g)^2 \in L(I)$, so $f + g$ is square Lebesgue integrable on $I$. $\blacksquare$
| Theorem 2 (Homogeneity): Let $f$ be square Lebesgue integrable functions on $I$ and let $c \in \mathbb{R}$. Then $cf$ is square Lebesgue integrable on $I$. |
- Proof: Let $f$ be square Lebesgue integrable on $I$ and let $c \in \mathbb{R}$. Since $f$ is square Lebesgue integrable on $I$ we have that $f \in M(I)$ and $f^2 \in L(I)$.
- Since $f \in M(I)$ there exists a sequence of step functions $(f_n(x))_{n=1}^{\infty}$ which converges to $f$ almost everywhere on $I$. So for any $c \in \mathbb{R}$, $(cf_n(x))_{n=1}^{\infty}$ is a sequence of step functions that converges to $cf$ almost everywhere on $I$. So $cf \in M(I)$.
- Now since $f^2 \in L(I)$ we have that $(cf)^2 = c^2f^2 \in L(I)$ by the linearity of the Lebesgue integral since $c^2$ is just a constant.
- Since $cf \in M(I)$ and $(cf)^2 \in L(I)$ we have that $cf$ is square Lebesgue integrable on $I$. $\blacksquare$