[proofplan] We prove the Lipschitz estimate directly from the definition. Fix two arbitrary points of the [metric space](/page/Metric%20Space), rewrite the difference of the pointwise sum as the sum of the two individual differences, and then apply the triangle inequality in the [normed vector space](/page/Normed%20Vector%20Space). The assumed Lipschitz bounds for $f$ and $g$ give the desired constant after combining the two terms. [/proofplan]

[step:Rewrite the difference of the pointwise sum] Define the pointwise sum map \begin{align*} h:X&\to V \end{align*} by \begin{align*} h(x)=f(x)+g(x). \end{align*} Let $x,y\in X$ be arbitrary. Since addition and subtraction in $V$ are the [vector space](/page/Vector%20Space) operations, we have \begin{align*} h(x)-h(y)=(f(x)+g(x))-(f(y)+g(y)). \end{align*} Rearranging terms in the vector space $V$ gives \begin{align*} h(x)-h(y)=(f(x)-f(y))+(g(x)-g(y)). \end{align*} [/step]

[step:Apply the norm triangle inequality and the two Lipschitz bounds]Taking the norm $\|\cdot\|_V$ and applying the triangle inequality in the normed vector space $V$, we obtain \begin{align*} \|h(x)-h(y)\|_V\le \|f(x)-f(y)\|_V+\|g(x)-g(y)\|_V. \end{align*} Because $f$ is $L$-Lipschitz, the first term satisfies \begin{align*} \|f(x)-f(y)\|_V\le Ld_X(x,y). \end{align*} Because $g$ is $M$-Lipschitz, the second term satisfies \begin{align*} \|g(x)-g(y)\|_V\le Md_X(x,y). \end{align*} Combining these estimates gives \begin{align*} \|h(x)-h(y)\|_V\le Ld_X(x,y)+Md_X(x,y). \end{align*} Factoring the common nonnegative quantity $d_X(x,y)$ yields \begin{align*} \|h(x)-h(y)\|_V\le (L+M)d_X(x,y). \end{align*}[/step]

[guided]The goal is to prove that the pointwise sum is Lipschitz with constant $L+M$. We therefore introduce the sum as an explicit map \begin{align*} h:X&\to V \end{align*} by \begin{align*} h(x)=f(x)+g(x). \end{align*} To check the Lipschitz condition, take arbitrary points $x,y\in X$. We must estimate $\|h(x)-h(y)\|_V$ in terms of $d_X(x,y)$. Using the definition of $h$, we first expand the difference: \begin{align*} h(x)-h(y)=(f(x)+g(x))-(f(y)+g(y)). \end{align*} The codomain $V$ is a vector space, so subtraction distributes over addition and the terms may be regrouped as \begin{align*} h(x)-h(y)=(f(x)-f(y))+(g(x)-g(y)). \end{align*} This decomposition is the key point: it separates the part controlled by the Lipschitz constant of $f$ from the part controlled by the Lipschitz constant of $g$. Now apply the triangle inequality for the norm $\|\cdot\|_V$ to the two vectors $f(x)-f(y)$ and $g(x)-g(y)$ in $V$: \begin{align*} \|h(x)-h(y)\|_V\le \|f(x)-f(y)\|_V+\|g(x)-g(y)\|_V. \end{align*} The hypotheses state exactly that $f$ is $L$-Lipschitz and $g$ is $M$-Lipschitz, meaning that for every pair of points in $X$, \begin{align*} \|f(x)-f(y)\|_V\le Ld_X(x,y) \end{align*} and \begin{align*} \|g(x)-g(y)\|_V\le Md_X(x,y). \end{align*} Substituting these two estimates into the triangle-inequality bound gives \begin{align*} \|h(x)-h(y)\|_V\le Ld_X(x,y)+Md_X(x,y). \end{align*} Finally, factor out $d_X(x,y)$: \begin{align*} \|h(x)-h(y)\|_V\le (L+M)d_X(x,y). \end{align*} Thus $h=f+g$ satisfies the Lipschitz estimate with constant $L+M$ for the arbitrary pair $x,y\in X$.[/guided]

[step:Conclude the Lipschitz estimate for all pairs of points] The points $x,y\in X$ were arbitrary, and the estimate \begin{align*} \|(f+g)(x)-(f+g)(y)\|_V\le (L+M)d_X(x,y) \end{align*} holds for every such pair. Hence $f+g:X\to V$ is $(L+M)$-Lipschitz. [/step]