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# Inequality in intergation (TIFR 2014 problem 6)

Question:

Let $$f:[0,1] \to \mathbb{R}$$ be a continuous function. Which of the following statement is always true?

A. $$\int_{0}^{1} f^2(x)dx = (\int_{0}^{1}f(x)dx)^2$$

B.  $$\int_{0}^{1} f^2(x)dx \le (\int_{0}^{1}|f(x)|dx)^2$$

C.  $$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$

D.  $$\int_{0}^{1} f^2(x)dx < (\int_{0}^{1}f(x)dx)^2$$

Discussion:

We first recall the Cauchy-Schwartz inequality for an inner product space $$V$$ and two vectors $$a,b\in V$$

$$<a,b> \le ||a||||b||$$.

Here, we also remember the fact that $$C[0,1]$$ (the set of all continuous real (/complex) valued functions on [0,1] ) forms an inner product space with respect to the inner product

$$<f,g>= \int_{0}^{1} f(x)g(x)dx$$ (We are only taking real valued functions so we neglect the conjugation…)

We want to apply this inequality to suitable functions so that we get some inequality from the options above.

Let’s try $$|f|$$ and $$1$$ (the constant function).

We get: $$\int_{0}^{1} |f(x)|1dx \le (\int_{0}^{1} |f(x)|^2dx)^{1/2} (\int_{0}^{1} 1^2dx)^{1/2}$$

Squaring, we get

$$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$.

So option C is correct.

October 21, 2017