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 $H$?
Best Answer
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.)