Algebraic Proof that a Disk is Convex

After searching on Google for a while, I cannot seem to find an algebraic proof that a disk is a convex set. Intuitively, this seems obvious: if you take any two points $x, y$ in a disk, then the line from $x$ to $y$ is clearly contained in the disk.

More specifically, let $D \subset \mathbb{R}^2$ be an origin-centered disk of radius $r$. Let $x = (x_1, x_2), y = (y_1, y_2) \in C$. We wish to show that for any $\lambda \in [0, 1]$, the point $z$ given by $$z = \lambda x + (1 - \lambda) y = (\lambda x_1 + (1 - \lambda y_1), \lambda x_2 + (1 - \lambda) y_2)$$ also lies in $D$. By definition, we have $\|x\|_2^2 \leq r^2$ and $\|y\|_2^2 \leq r^2$. Using this information, I tried to show that $\|z\|_2^2 \leq r^2$ as follows: \begin{align*} (\lambda x_1 + (1 - \lambda y_1))^2 + (\lambda x_2 + (1 - \lambda) y_2)^2 &= \lambda^2 x_1^2 + 2 \lambda (1 - \lambda) x_1 y_1 + (1 - \lambda)^2 y_1^2 + \\ &\phantom{50} \lambda_2 x_2^2 + 2 \lambda (1 - \lambda) x_2 y_2 + (1 - \lambda)^2 y_2^2 \\ &= \lambda^2 (x_1^2 + x_2^2) + 2 \lambda (1 - \lambda) (x_1 y_1 + x_2 y_2) + (1 - \lambda)^2 (y_1^2 + y_2^2). \end{align*} Unfortunately, I'm stuck at this point -- I can't see a clean way to factor this expression and bound it above by $r^2$. I feel like I am missing obvious, but at this time I cannot determine what that is. Is there a clean proof of this fact that follows from this reasoning?

Edit: Solution using Cauchy-Schwarz Inequality

Even though I used the notation for norms, it didn't occur to me to just use the triangle inequality, as Dustan Levenstein suggested. The solution would then proceed as follows: \begin{align*} \| \lambda x + (1 - \lambda) y \|_2 &\leq \lambda \| x \|_2 + (1 - \lambda) \| y \|_2 \\ &\leq \lambda r + (1 - \lambda) r = r. \end{align*} It follows that $\| z \|_2^2 \leq r^2$, as desired. I'm still going to leave the question up, in case there's a solution that does not rely on the Cauchy-Schwarz inequality.

• It's a consequence of the triangle inequality for normed vector spaces. Commented Mar 11, 2014 at 4:09
• @DustanLevenstein I didn't think about it in terms of norms even though I used that notation; thanks. I've edited the question to include a solution that uses the triangle inequality. Commented Mar 11, 2014 at 4:10
• @void-pointer you don't even need an inner product. Just apply the triangle inequality directly to $\|\lambda x + (1-\lambda)y\|$. Commented Mar 11, 2014 at 4:15
• @DustanLevenstein Yes, I've just edited the question. The answer follows easily from the triangle inequality. Commented Mar 11, 2014 at 4:18

Equation of a disk: $$x^2+y^2≤r^2$$ (r is the radius).

Let’s say 2 points in the disk are $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Now we have to show that any point on the line segment joined by these 2 points i.e. $$(tx_1+(1-t)x_2, ty_1+(1-t)y_2)$$ also satisfies the disk equation, where $$1 \leq t \leq 0$$.

$$(tx_1+(1-t)x_2)^2 + (ty_1+(1-t)y_2)^2$$

$$= t^2 x_1^2+t^2 y_1^2+(1-t)^2 x_2^2+(1-t)^2 y_2^2 + 2t(1-t)[x_1x_2+y_1y_2]$$

$$= t^2(x_1^2+y_1^2)+(1-t)^2(x_2^2+y_2^2) + t(1-t)(2x_1x_2 + 2y_1y_2)$$

$$\leq t^2 r^2+ (1-t)^2 r^2 + t(1-t)(x_1^2+x_2^2+y_1^2+y_2^2)$$, because $$2xy≤x^2+y^2$$,prove it yourself.

$$\leq t^2r^2+ (1-t)^2r^2 + t(1-t)(r^2+r^2) = r^2 [t^2+(1-t)^2 + 2t(1-t)]$$

$$= r^2$$

Hence proven disk is convex.

• MathJax tutorial for typing math. Commented Jun 3, 2018 at 12:14
• I've made the changes in MathJax. Also Snehasish, thanks for the solution. Commented Apr 4, 2020 at 12:15
• Why 2𝑥𝑦≤𝑥2+𝑦2? @AnirbanSaha
– MAC
Commented Jan 30 at 15:16
• @MAC $0 \leq x^2 - 2xy + y^2 = (x-y)^2$ Commented Feb 16 at 22:54

It's no extra work to prove it for an ellipsoid.

Given a positive (or non negative) definite bilinear form on a denoted $(x,y)$, the convexity of $f(x)=(x,x)$ is, after expanding all terms and cancelling the non-negative scalar factor of $\lambda(1-\lambda)$, the statement that

$$(a,a) + (b,b) \geq (a,b) + (b,a)$$

which is the expansion of $(a-b,a-b) \geq 0$.

I had thought it would be equivalent to Cauchy Schwarz but it looks a little bit weaker.