# Convexity of a functional - most straightforward way

Consider the functional $$H(x,y,y'):=\int\limits_0^\pi \left[2\sin(x)y(x)+y'(x)^2-\lambda y(x) \right]\mathbb{d}x$$

We can find that the stationary function for this functional is

$$y_0(x) = \frac{8x}{\pi^3}(x-\pi)-\sin(x)$$

Now I need to find out whether this is a minimizing function using the concept of convexity. We can find that if the functional $H$ is convex then $y_0(x)$ is the unique minimizing function for $H$.

We can use the Hessian matrix determinant, which is $0$, to show that the integrand of $H$ is convex, which implies that $H$ is at least strongly convex [please correct me if I'm wrong].

Is there a more straightforward way to determine convexity of $H$?

• Please check the definition of $H$. The variable $x$ is not a free variable of the right-hand side. – gerw Oct 24 '16 at 6:52
• @gerw Sorry, can you please clarify that? – sequence Oct 24 '16 at 19:16
• Your right-hand side does not depend on $x$. The left-hand side depends on $x$. – gerw Oct 24 '16 at 21:07
• How about $2\sin(x)$? – sequence Oct 25 '16 at 7:00
• Let me give a simple example: Take $H = \int_0^\pi \sin(2 \, x) \, \mathrm{d}x$. This does not depend on $x$, although $\sin(2 \, x)$ depends on $x$. In fact, you have $H = 0$. – gerw Oct 25 '16 at 9:26

There is a more direct way using the definition of convexity via linear combinations. In fact, your functional $H(x,y,y')$ can be defined as a functional $F(y)$ which maps, say, $C^1[0,\pi]$ to $\mathbb{R}$. (This also relates with the comment of gerw, by the way).
So, we have $$F(y) := \int\limits_0^\pi \left[2\sin(x)y(x)+y'(x)^2-\lambda y(x) \right]\mathbb{d}x.$$ By the definition of convexity, we have to show that for any $y_1, y_2 \in C^1[0,\pi]$ and any $t \in [0,1]$, it holds $$F(ty_1 + (1-t)y_2) \leq tF(y_1) + (1-t)F(y_2).$$ However, this fact is somehow obvious, since $y$ is included linearly into $F$, and $y'$ is included quadratically with proper sign. That is, each summand is convex and so is the whole $F$. (Nevertheless, you can check it directly by definition.)
• So, if $y'(x)$ weren't squared, we'd have $F(y,y')$? – sequence Oct 25 '16 at 17:17
• @sequence No) we have $F(y)$ in any case. Roughly, functional - is a map which maps a set of functions to numbers. For example, let us fix some function $y$ and you know that $y$ is differentiable. To define a functional, you can make with $y$ (almost) whatever you want (integrate it, differentiate, take square roots, multiply by some other functions, etc.), but at the end you must get a number. A definite integral in your example - is a number (for each fixed $y$ (and hence $y'$ is also fixed)). So, you get a functional, which maps continuously differentiable functions to numbers. – Voliar Oct 25 '16 at 21:05