# How to prove that the sum of convex sets is convex?

Below we have the definition of a convex set.I want to prove that sum of convex sets is a convex set.using definition bellow i take two points from each set $$x'_1,x'_2\in S1$$ and $$x''_1, x''_2 \in S2$$. For each set we have the following expression

$$\lambda'x'_1+(1-\lambda')x'_2 \in S1 \\ \lambda''x''_1+(1-\lambda'')x''_2 \in S2$$

If i want to show that their sum is convex too,i need to arrange it in the $$\lambda x+(1-\lambda)x \in S1$$ structure structure. How can i do it? Thanks

Update:I have seen solutions to these proof as shown bellow. the main problem with this proof is that they share the same lambda unlike what i tried to have separated lambdas. Why is that?  • Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. Apr 6, 2021 at 16:47
• Hello Saulspatz, i have updated the original post. Apr 6, 2021 at 17:12
• In your proof, consider two points $a_1$ and $b_1$. If they both lie in $X$ or $Y$ you are done as $X,Y\subset X+Y$, and X, Y are convex. Consider one of them lying in $X$ and the other in $Y$, then you just need 1 lambda. See here: math.stackexchange.com/questions/2410875/… Apr 6, 2021 at 17:24
• Hello Rahul,suppose we have two points $a_x,b_x \in S_1 + S_2$ specifically $a_1,a_2 \in S_1$ and $b_1,b_2 \in S_2$, so we need to take two points where each one is a linear combination of the point from S1 and S2.$a_x=a_1+a_2$ $a_y=b_1+b_2$ so we get $\lambda a_x +(1-\lambda) a_y=(\lambda a_1+(1-\lambda)a_2)+(\lambda b_1+(1-\lambda)b_2) \\$ So we have a sum of two points each one from convex field,how does it made the sum convex too?basicly we are using what we need to prove in order to solve the issue. we get sum of two point from convex fields how exactly it solves the issue?Thanks Apr 6, 2021 at 18:12

Given: $$S_1$$ and $$S_2$$ are convex sets.
To Show: $$S_1+S_2$$ is also convex. Note that $$S_1+S_2=\{s;s=s_1+s_2 \text{ such that } s_1\in S_1\text{and }s_2\in S_2\}$$
Let $$s,t\in S_1+S_2$$. Fix some $$\lambda\in[0,1]$$. (I guess here is where the doubt is. As we have to show convexity of the set $$S_1+S_2$$, we need not see them as separate entities, we only need to keep in mind the form of the components in that set).
Then $$s=s_1+s_2, t=t_1+t_2$$ such that $$s_1,t_1\in S_1$$ and $$s_2,t_2\in S_2$$.
$$\lambda s+(1-\lambda)t=\lambda(s_1+s_2)+(1-\lambda)(t_1+t_2)=(\lambda s_1+(1-\lambda)t_1)+(\lambda s_2+(1-\lambda)t_2)\in S_1+S_2$$. (As both sets $$S_1$$ and $$S_2$$ are convex). Thus, $$S_1+S_2$$ is also convex.
• Hello Cherryblossoms,i agree with everything ,I am having a logical problem to understand the last line in your explanation. why if the sum is assembled from points in convex fields,then their sum is convex too? We are using what we need to prove to solve the issue,we didnt prove that its convex. we just proved that out two points $s,t$ have in the middle a point which could be a sum of two ponts from convex field,thats it.hope you understand my problem. "(As both sets $S_1$ and $S_2$ are convex). Thus, $S_1+S_2$ is also convex." Apr 6, 2021 at 19:01
• @rocko445 The proof is correct. It shows that any convex linear combination of $2$ points from $S+T$ is an element of $S+T$. That's what it means for $S+T$ to be convex. I don't know where you think it uses circular reasoning; it doesn't. Apr 6, 2021 at 19:41
• @rocko445 If the sum is assembled from points in convex fields, their sum is also convex: This is because the elements in $S_1+S_2$ have that form. I would suggest that you try seeing $S_1+S_2$ as a set, call it $X$. Probably you are having a problem with the way the set is written. Then apply the definition of convexity. We did prove that $X$ is convex and we used the fact that $S_1$ and $S_2$ are convex, as that is given. As for your last line, any point which lies in the middle of $s$ and $t$ will be a sum of points from the convex fields. This is exactly what you need to show. Apr 7, 2021 at 4:53