If $\alpha,\beta \in BV[a,b]$, prove that $|a| \in BV[a,b]$ and $\min(\alpha,\beta),\max(\alpha,\beta) \in BV[a,b]$.

Question

Question:

If $\alpha,\beta \in BV[a,b]$, prove that $|a| \in BV[a,b]$ and $\min(\alpha,\beta),\max(\alpha,\beta) \in BV[a,b]$.

Attempt:

Since $\alpha \in BV[a,b]$, $$ \sum_{i=1}^n \left| \alpha(x_i) - \alpha(x_{i-1}) \right| \leq V_a^b \alpha. $$ By the triangle inequality, we have $$ \sum_{i=1}^n \left| |\alpha(x_i)| - |\alpha(x_{i-1})| \right| \leq \sum_{i=1}^n \left| \alpha(x_i) - \alpha(x_{i-1}) \right| \leq V_a^b \alpha $$ so $|\alpha| \in BV[a,b]$. For the second part, we have \begin{align*} \sum_{i=1}^n \left| \min( \alpha_i - \alpha_{i-1},\beta_i - \beta_{i-1}) \right| &< \sum_{i=1}^n \left( |\alpha(x_i) - \alpha(x_{i-1})| + |\beta(x_i) - \beta(x_{i-1})| \right) \\ &\leq V_a^b \alpha + V_a^b \beta. \end{align*} and similarly \begin{align*} \sum_{i=1}^n \left| \max( \alpha_i - \alpha_{i-1},\beta_i - \beta_{i-1}) \right| &< \sum_{i=1}^n \left( |\alpha(x_i) - \alpha(x_{i-1})| + |\beta(x_i) - \beta(x_{i-1})| \right) \\ &\leq V_a^b \alpha + V_a^b \beta. \end{align*} Hence, $\min(\alpha,\beta),\max(\alpha,\beta) \in BV[a,b]$.

Is this proof correct?

Answer 1

The proof is correct. Another, lazier approach: after proving that $|\alpha|\in BV$ use the identities $$ \max(\alpha,\beta)= \frac12 (\alpha+\beta+|\alpha-\beta|) $$ $$ \min(\alpha,\beta)= \frac12 (\alpha+\beta-|\alpha-\beta|) $$ together with the fact that BV is a linear space.