Integral and inequality

Let f:[0,1] \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} \int_0^1 f(x) \, \mathrm{d}x = \kappa =\int_0^1 x f(x) \, \mathrm{d}x \end{equation*}

Prove that

    \[\int_0^1 f^2(x) \, \mathrm{d}x \geq 4\kappa^2\]

Solution

We note that

    \begin{align*} \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x &= \int_{0}^{1} f^2(x) \, \mathrm{d}x -12 \kappa \int_{0}^{1} x f(x) \, \mathrm{d}x + 36 \kappa^2 \int_{0}^{1} x^2 \, \mathrm{d}x \\ &=\int_{0}^{1} f^2(x) \, \mathrm{d}x - 12 \kappa \int_{0}^{1} f(x) \, \mathrm{d}x + \frac{36\kappa^2}{3} \\ &=\int_{0}^{1} f^2 (x) \, \mathrm{d}x - 12 \kappa^2 + 12 \kappa^2\\ &= \int_{0}^{1} f^2(x) \, \mathrm{d}x \end{align*}

On the other hand by Cauchy – Schwartz we have that

    \begin{align*} \int_{0}^{1} f^2(x) \, \mathrm{d}x &= \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x \\ &\geq \left [ \int_{0}^{1} \left ( f(x) - 6 \kappa x \right ) \, \mathrm{d}x \right ]^2 \\ &=\left ( \kappa - 3 \kappa \right )^2 \\ &= 4\kappa^2 \end{align*}

The result follows.

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Determinant of block matrices

Given a real n \times n matrix A show that \begin{vmatrix} A &A^2 \\ A^3 & A^4 \end{vmatrix} =0.

Solution

We simply note that

    \[\begin{pmatrix} A &\mathbb{O} \\ A^3 & \mathbb{O} \end{pmatrix} \cdot \begin{pmatrix} \mathbb{I} & A\\ \mathbb{O} & \mathbb{O} \end{pmatrix} = \begin{pmatrix} A &A^2 \\ A^3 & A^4 \end{pmatrix}\]

The rest is … history!

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Limit of a sum

Let a_n be a real sequence such that a_n>0 , \liminf_{n \rightarrow +\infty} a_n=1 , \limsup a_n =2 and \displaystyle \lim_{n \rightarrow +\infty} \sqrt[n] {\prod_{k=1}^n{a_k}}=1. Prove that

    \[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} a_k =1\]

Solution

Lemma: Let a_n be a bounded sequence of, say, complex numbers and let \ell be another complex number.

    \[\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k=\ell \Leftrightarrow \text{D-lim } a_n=\ell\]

where the latter means “convergence on a set of density 1 ” i.e. there exists a subsequence n_1,n_2,\dotsc such that \lim \limits_{k \to +\infty} a_{n_k}=\ell  and n_k is dense i.e.

    \[\lim \limits_{n \to +\infty} \frac{\vert \{1,2,\dotsc,n\} \cap \{n_1,n_2,\dotsc,\}\vert}{n}=1\]

Proof: The proof is omitted because it is too technical.

Hence,

    \begin{align*} \lim_{n \rightarrow +\infty} \sqrt[n]{\prod_{k=1}^n a_k}=1 &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n \log a_k=0 \\ &\Leftrightarrow \text{D-lim } \log a_n=0 \\ &\Leftrightarrow \text{D-lim } a_n=1 \\ &\Leftrightarrow \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^n a_k=0 \end{align*}

All we need is that the logarithm is continuous and that a_n as well as \log a_n are bounded sequences.

Note: We can simplify the conditions to a_n>0, \liminf \limits_{n \rightarrow +\infty} a_n>0 and \limsup \limits_{n \rightarrow +\infty} a_n< +\infty.

 

The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.

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Rational number

Prove that there exist infinite many positive real numbers x such that the sum

    \[\mathcal{S}(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n+x}\]

is rational.

Solution

It suffices to prove that \mathcal{S} is continuous. Indeed,

    \begin{align*} \left | \mathcal{S}(x) - \mathcal{S}(y) \right | &=\left | \sum_{n=1}^{\infty} (-1)^{n-1} \left ( \frac{1}{n+x} - \frac{1}{n+y} \right ) \right | \\ &\leq \sum_{n=1}^{\infty} \frac{\left | x-y \right |}{\left ( n+x \right )\left ( n+y \right )} \\ &\leq \left | x-y \right | \sum_{n=1}^{\infty} \frac{1}{n^2} \\ &\leq 4 \left | x-y \right | \end{align*}

Thus \mathcal{S} is continuous. Since \mathcal{S} is non constant , it follows that \mathcal{S}((0, +\infty)) is an interval. The result follows.

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On the zero function

Let f:\mathbb{R} \rightarrow \mathbb{R} such that

(1)   \begin{equation*} \left | f(x) \right | \leq \left | \int_{0}^{x} f(t) \, \mathrm{d}t \right | \quad \quad \text{forall} \;\;\; x\in \mathbb{R} \end{equation*}

Prove that f(x)=0 \; , \; x \in \mathbb{R}.

Solution

Basically, it’s this exercise. Can you see the reason why?

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