Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex.

Question

Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$ \begin{align} \int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0 \end{align}

I am completely stumped. What I have tried to do is return the query for $k=1$, and for that value of $k$ try to write the integral from $0$ to $2\pi$ as a sum of four integrals from $0$ to $\dfrac{\pi}{2}$ and use the the theorem for first derivative monotony. No luck so far.

Any help would be much appreciated.

Edit 1: I saw the link here about a similarly asked topic. However, this process gets the general case as I perceive it and I am really supposed to use the method described above. I will try this and come back with a definitive answer.

Edit 2: I cannot use the $f''(x) \geq 0$ argument due to the simple fact that I have not been formally taught it as part of the class.

Edit 3: Final proof, with thanks to the contributors below.

Let's start by setting $A = \frac{\pi}{2}$ and $B = \frac{3\pi}{2}$. It follows from basic trigonometry that: \begin{align} &\cos x \geq 0, \ x \in [A,B] \ \text{and}\\ &\cos x \leq 0, \ x \in [0, A] \cap [B, 2\pi]. \end{align}

And we also set $L$ to be the line segment such that $L(A) = f(A), \ L(B) = f(B)$. We will prove a basic property of said line in regards to the convex function $f$.

1. I can take for granted that (we proved this in class) \begin{align} L(x) = \dfrac{x-A}{B-A}f(B) + \dfrac{B-x}{B-A}f(A) \end{align} so for $x \in [A,B]$ there exists $\lambda \in [0,1]$ such that: $x = \lambda A + (1-\lambda) B$. Taking the aforementioned expression and replacing it on $L(x)$ we get (I omit trivial algebra) \begin{align} L(x) = (1-\lambda) f(B) + \lambda f(A). \end{align} Since $f$ is convex, we can write \begin{align} &f(\lambda A + (1-\lambda) B) \leq \lambda f(A) + (1-\lambda) f(B)\\ \implies &f(x) \leq L(x), \ \forall \ x \in [A,B] \ \text{and} \ \lambda \in [0,1]. \end{align} $\blacksquare$

2. We assume that there exists $x \in [0,A]: \ f(x) < L(x)$. Then there exists $A \in [x, B] \ \text{and} \ \lambda \in [0,1]: \ A = \lambda x + (1-\lambda) B$. Then \begin{align} &L(A) = \dfrac{A-x}{B-x}f(B) + \dfrac{B-A}{B-x}f(x), \ A \in [x,B]\\ \implies &L(A) = (1-\lambda)L(B)+\lambda L(x)\\ \implies &L(A) = (1-\lambda)f(B) + \lambda L(x). \end{align} Then assuming that $L(x) > f(x)$ we get \begin{align} L(A) > (1-\lambda) f(B) + \lambda f(x) \geq f(A), \ \text{assuming convexity}. \end{align} Because $L(A) = f(A)$ the above inequality becomes $L(A) > f(A)$ which is a contradiction.

$\blacksquare$

In the same spirit, for $x \in [B, 2\pi]$ doing the exact same replacements and applying the exact same principles we also get \begin{align} &f(\lambda B + (1-\lambda) 2 \pi) \leq \lambda f(B) + (1-\lambda) f(2\pi)\\ \implies &f(x) \leq L(x). \end{align}

$\blacksquare$

We then set $g(x) = f(x) - L(x)$ and from (1) and (2) above it is safe to assume that $\cos x$ and $g(x)$ will have the same sign in the whole domain, that is: \begin{align} &g(x) \geq 0, \ \text{where} \ \cos x \geq 0\\ &g(x) \leq 0, \ \text{where} \ \cos x \geq 0. \end{align}

3. We have \begin{align} \int_0^{2\pi} g(x) \cos x dx = \int_0^{\frac{\pi}{2}} g(x) \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} g(x) \cos x dx + \int_{\frac{3\pi}{2}}^{2\pi}g(x) \cos x dx \geq 0, \end{align} because \begin{align} &\text{in} \left[0, \frac{\pi}{2} \right], \ \cos x \geq 0 \implies g(x) \geq 0\\ &\text{in} \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right], \ \cos x \leq 0 \implies g(x) \leq 0\\ &\text{in} \left[\frac{3 \pi}{2}, 2 \pi \right], \ \cos x \geq 0 \implies g(x) \geq 0.\\ \end{align}

4. We have that \begin{align} &\int_0^{2 \pi}\cos x dx = 0 \ \text{trivial}\\ &\int_0^{2 \pi}x \cos x dx = \int_0^{2 \pi}x (\sin x)' dx = \left[ x \sin x \right]_0^{2\pi} - \int_0^{2 \pi} \sin x dx = 0. \end{align}

It then follows that \begin{align} \int_0^{2 \pi} f(x) \cos x dx = \int_0^{2 \pi} g(x) \cos x dx \geq 0 \end{align} which was previously proven. For the case $k=1$, the proof is over.

For $k>1$, we have: \begin{align} \int_0^{2\pi}f(x) \cos (kx) dx = \sum_{i=0}^{k-1} \int_{\frac{2\pi i}{k}}^{\frac{2\pi (i+1)}{k}} f(x) \cos (kx) dx. \end{align} We perform the change of variable \begin{align} x = \dfrac{y+2\pi i}{k}\\ \implies \begin{cases}dx = \dfrac{1}{k}dy\\ x = \dfrac{2\pi i}{k} \to y=0\\ x = \dfrac{2\pi (i+1)}{k} \to y = 2\pi \end{cases}. \end{align} So the above sum becomes \begin{align} \sum_{i=0}^{k-1} \dfrac{1}{k} \int_{0}^{2\pi}f \left(\dfrac{y+2 \pi i}{k} \right) \cos (y+2\pi i)dy. \end{align} Because $f$ is convex in $[0, 2\pi]$ there exists $\lambda \in [0,1]$ such that: \begin{align} \theta f\left(\frac{y_1 + 2\pi i}{k}\right) + (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right) &\ge f\left(\theta\frac{y_1 + 2\pi i}{k} + (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\ &= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right). \end{align}

Having performed the change of variables: \begin{align} x_1 = \dfrac{y_1 + 2\pi i}{k} \ \text{and} \ x_2 = \dfrac{y_2 + 2\pi i}{k}. \end{align}

So $f \left( \dfrac{y + 2\pi i}{k}\right)$ convex on $[0, 2\pi]$. Using the result from $k=1$ we have \begin{align} \int_0^{2\pi} f \left( \dfrac{y + 2\pi i}{k} \right)\cos y dy \geq 0. \end{align}

$\blacksquare$

Answer 1

I will try to give you some hints that can give you a proof using essentially only the definition of convexity.

Start by the case $k=1$. Denote $A=\pi/2$, $B=3\pi/2$, so that $\cos x\geq 0$ in $[A, B]$ and $\cos x\leq 0$ in $[0,A]\cap[B,2\pi]$. Let $L$ be the straight line such that $L(A)=f(A)$ and $L(B)=f(B)$.

1. Show that $f(x)\leq L(x)$ for all $x\in[A, B]$. (Hint: just apply the definition of convexity...)

2. Show that $f(x)\geq L(x)$ in $[0,A]\cap[B,2\pi]$. (Hint: assume that there is a point $x$ such that this is not true and try to find the contradiction...)

Now we can consider the function $g = f-L$. From the points (1) and (2) you have that, on $[0,2\pi]$, $g\geq 0$ where $\cos\geq 0$ and $g\leq 0$ where $\cos\leq 0$.

3. Conclude that $\int_0^{2\pi}g(x)\cos x dx\geq 0$.

4. Show that $\int_0^{2\pi}f(x)\cos x dx=\int_0^{2\pi}g(x)\cos x dx$. (Hint: first show that $\int_0^{2\pi}\cos x dx=0$ and $\int_0^{2\pi}x\cos x dx=0$...)

Now you need to consider the generic case $k\geq 1$. However this follows from the case $k=1$. The idea is that we can split $[0, 2\pi]$ in $k$ intervals of length $2\pi/k$. Now let $[a, b]$ be one of this intervals, that is for some integer $0\leq m<k$ you have $ak=2m\pi$ and $bk=2(m+1)\pi$. We need to show that $\int_a^b \cos(kx)f(x)dx\geq 0$.

5. Show that $\int_a^b \cos(kx)f(x)dx = \frac{1}{k}\int_0^{2\pi}f\left(a+x/k\right)\cos(x)dx$. (Hint: this is just a change of variable...)

6. Show that $x\mapsto f\left(a+x/k\right)$ is convex and conclude that $\int_a^b \cos(kx)f(x)dx\geq 0$.

7. Show that $\int_0^{2\pi}f(x)\cos(kx)dx\geq 0$.

I hope that this will help you.

Answer 2

The proof using twice integration by parts requires differentiability of $f.$ Instead we can use summation by parts as follows. The integral in question is the limit of Riemann sums $$S_n:={1\over n}\sum_{j=1}^n f\left ({2\pi j\over n}\right )\cos {2\pi k j\over n}$$ It suffices to show that $\lim S_n\ge 0.$

Observe that $$2\sin{\pi k \over n}\cos{2\pi kj\over n}=\sin {2\pi k(j+{1\over 2}) \over n}-\sin {2\pi k(j-{1\over 2}) \over n}$$ Therefore denoting $a_j=f\left ({2\pi j\over n}\right )$ and $b_j=\sin {2\pi k(j+{1\over 2}) \over n}$ leads to $$2n\sin\ {\pi k \over n}\ S_n=\sum_{j=1}^n a_j(b_j-b_{j-1}) =\sum_{j=0}^{n-1} (a_j-a_{j+1})b_j +a_{n}b_n-a_0b_0\\ =\sum_{j=0}^{n-1} (a_j-a_{j+1})b_j+[f(1)-f(0)]\,\sin{\pi k\over n} $$ We have $$ 2\sin {\pi k \over n}\cdot b_j=\left [1-\cos{2\pi k (j+1)\over n}\right ]-\left [1-\cos{2\pi k j\over n}\right ]$$ Hence denoting $\Delta a_j=a_j-a_{j+1},$ $c_j=1-\cos{2\pi k j\over n}$ and observing that $c_0=c_n=0$ gives $$ 4n\sin^2{\pi k\over n} \ S_n= \sum_{j=0}^{n-1}\Delta a_j\,[c_{j+1}-c_{j}]+ 2[f(1)-f(0)]\,\sin^2{\pi k\over n}\\ = \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+\Delta a_{n-1}\,c_n-\Delta a_0\,c_0 + 2[f(1)-f(0)]\,\sin^2{\pi k\over n}\\ = \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+ 2[f(1)-f(0)]\,\sin^2{\pi k\over n}$$ Summarizing $$S_n={1\over 4n\sin^2{\pi k\over n}} \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+{1\over 2n}[f(1)-f(0)]$$ The sum is nonnegative as $c_j\ge 0$ and $\Delta a_{j-1}-\Delta a_j=a_{j-1}+a_{j+1}-2a_j\ge 0$, by the convexity of $f$. Hence $\lim S_n$ is nonnegative.

Answer 3

As demonstrated in the other answers, it suffices to prove the statement for $k=1$, that is $$ I = \int_{0}^{2\pi}f(x) \cos (x) \, dx \ge 0 \, . $$ We can split the integral into four parts over the intervals $[0, \pi/2]$, $\pi/2, \pi]$, $[\pi , 3\pi/2]$ and $[3\pi/2, 2 \pi|$ and substitute $x = \pi - y$, $x=\pi + y$, $x =2\pi -y$ in the second, third, and fourth integral, respectively. This gives $$ I = \int_0^{\pi/2} \bigl( f(x) - f(\pi -x) - f(\pi + x) + f(2\pi - x)\bigr) \cos(x) \, dx \, . $$ So order to prove $I \ge 0$ it suffices to show that $$ \tag{$*$} f(\pi -x) + f(\pi + x) \le f(x) + f(2\pi - x) $$ for $0 \le x \le \pi/2$.

And that follows directly from the convexity condition, applied to $x < \pi - x< 2 \pi-x$: $$ f(\pi - x) \le \frac{\pi}{2 \pi - 2x}f(x) + \frac{\pi - 2x}{2 \pi - 2x}f(2\pi - x) \, , $$ and to $x < \pi + x < 2 \pi-x$: $$ f(\pi + x) \le \frac{\pi-2x}{2 \pi - 2x}f(x) + \frac{\pi}{2 \pi - 2x}f(2\pi - x) \, . $$ Adding these two inequalities gives exactly $(*)$, and that finishes the proof.

Answer 4

Supplement to @ECL's answer:

1. For $k = 1$:

Let $$g(x) := f(x) - \frac{f(3\pi/2) - f(\pi/2)}{\pi}x - \frac{\frac{3\pi}{2}f(\pi/2) - \frac{\pi}{2}f(3\pi/2)}{\pi}.$$

We have $g(\pi/2) = g(3\pi/2) = 0$. Also, $g(x)$ is convex on $[0, 2\pi]$.

For $x \in [\pi/2, 3\pi/2]$, letting $t = \frac{3}{2} - \frac{x}{\pi} \in [0, 1]$, we have $x = \frac{\pi}{2} t + \frac{3\pi}{2}(1 - t)$ and $$g(x) \le t g(\pi/2) + (1 - t)g(3\pi/2) = 0.$$

For $x\in [0, \pi/2]$, using $\pi - x \in [\pi/2, 3\pi/2]$ and $g(x) + g(\pi - x) \ge 2g(\pi/2) = 0$, we have $$g(x) \ge -g(\pi - x) \ge 0.$$

For $x\in [3\pi/2, 2\pi]$, using $3\pi - x\in [\pi/2, 3\pi/2]$ and $g(x) + g(3\pi - x) \ge 2g(3\pi/2) = 0$, we have $$g(x) \ge -g(3\pi - x) \ge 0.$$

Using $\int_0^{2\pi} \cos x \mathrm{d} x = 0$ and $\int_0^{2\pi} x \cos x \mathrm{d} x = 0$, we have \begin{align*} &\int_0^{2\pi} f(x) \cos x \mathrm{d} x\\ =\,& \int_0^{2\pi} g(x)\cos x \mathrm{d} x\\ =\,& \int_0^{\pi/2} g(x)\cos x \mathrm{d} x + \int_{\pi/2}^{3\pi/2} g(x)\cos x \mathrm{d} x + \int_{3\pi/2}^{2\pi} g(x)\cos x \mathrm{d} x\\ \ge\,& 0. \end{align*}

$\phantom{2}$

2. For $k > 1$:

We have \begin{align*} &\int_0^{2\pi} f(x)\cos k x \mathrm{d} x\\ =\,& \sum_{i=0}^{k-1} \int_{2\pi i/k}^{2\pi(i + 1)/k} f(x)\cos k x \,\mathrm{d} x\\ =\,& \sum_{i=0}^{k-1} \frac{1}{k}\int_0^{2\pi} f\left(\frac{y + 2\pi i}{k}\right)\cos y\, \mathrm{d} y. \end{align*}

For any $y_1, y_2 \in [0, 2\pi]$ and $\theta \in [0, 1]$, we have \begin{align*} \theta f\left(\frac{y_1 + 2\pi i}{k}\right) + (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right) &\ge f\left(\theta\frac{y_1 + 2\pi i}{k} + (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\ &= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right). \end{align*} Thus, $f\left(\frac{y + 2\pi i}{k}\right)$ is convex on $[0, 2\pi]$.

Using the result for $k = 1$, we have $$\int_0^{2\pi} f\left(\frac{y + 2\pi i}{k}\right)\cos y \,\mathrm{d} y \ge 0.$$

The desired result follows.

Answer 5

Without loss of generality we may assume that $f$ is twice continuously differentiable on $[0,2\pi].$ Then $$ \int_0^{2\pi}f(x)\cos(kx)\,dx=\left. f(x)\frac {\sin(kx)} k\right |_0^{2\pi} -\int_0^{2\pi} f'(x)\frac {\sin(kx)} k\,dx= -\int_0^{2\pi} f'(x) \frac {\sin(kx)} k\,dx.$$ Next, again integraiting by parts, this equals $$\left. f'(x)\frac {\cos(kx)} k\right|_0^{2\pi} - \int_0^{2\pi}f''(x)\cos(kx)\,dx\ge$$ $$f'(2\pi)-f'(1)-\int_0^{2\pi}f''(x)\cdot1\,dx= $$ $$f'(2\pi)-f'(1)-\int_0^{2\pi} f''(x)\,dx=0 $$ since the second derivative of a convex function is nonnegative. The general case is obtained by approximation of $f(x)$ by a smooth function with precision $\epsilon$, making use of convolution, and then by taking the limit as $\epsilon\to 0+$.