Assume first that f and g are real-valued and not both identically zero. For any real parameter λ, consider the integral of the square \( \displaystyle \quad \left( f - \lambda\, g \right)^2 . \quad \) Expanding the square under the integral sign gives \[ \int_a^b \left( f(x) - \lambda\, g(x) \right)^2 {\text d}x = \int_a^b \left( f^2 (x) - 2\lambda\,f(x)\,g(x) + g^2 (x) \right) {\text d}x . \] By linearity of the integral, this becomes \[ \int_a^b f(x)^2\, {\text d}x\; -\; 2\lambda \int_a^b f(x)\,g(x)\, {\text d}x\; +\; \lambda ^2\int_a^b g(x)^2\, {\text d}x\; \geq \; 0. \] Introduce the abbreviations \[ A=\int _a^bg(x)^2\, {\text d}x,\quad B=\int _a^bf(x)g(x)\, {\text d}x,\quad C=\int _a^bf(x)^2\, {\text d}x. \] Then the inequality above reads \[ A\lambda ^2-2B\lambda +C\; \geq \; 0\quad \mathrm{for\ all\ }\lambda \in \mathbb{R}. \] This is a quadratic polynomial in λ. A real quadratic polynomial Aλ² - 2Bλ + C with A ≥ 0 is nonnegative for all λ if and only if its discriminant is nonpositive: \[ (-2B)^2-4AC\; \leq \; 0\quad \Longleftrightarrow \quad 4B^2-4AC\; \leq \; 0\quad \Longleftrightarrow \quad B^2\; \leq \; AC. \] Substituting back the definitions of A,B,C, we obtain \[ \left( \int _a^bf(x)g(x)\, {\text d}x\right) ^2\; \leq \; \left( \int _a^bf(x)^2\, {\text d}x\right) \left( \int _a^bg(x)^2\, {\text d}x\right) , \] which is precisely Bunyakovsky’s inequality.

Equality case: From the argument above, equality holds if and only if the discriminant is zero and the quadratic form \[ A\lambda ^2-2B\lambda +C \] has a unique real root λ₀. Since the integrand is nonnegative, the integral can vanish only if \[ f(x)-\lambda_0 g(x) = 0\quad \mathrm{for\ almost\ every\ }x\in [a,b], \] i.e., f and g are linearly dependent in 𝔏²[𝑎, b].

Bunyakovsky's inequality is naturally extended for inner-product-spaces (conceptual view). In the space 𝔏²[𝑎, b] of square–integrable functions, define a real inner product as \( \displaystyle \quad \langle f, g \rangle = \int f(x)\,g(x)\,{\text d}x . \quad \) Then Bunyakovsky’s inequality becomes exactly the Cauchy–Schwarz inequality in this inner product space: \[ |\langle f,g\rangle |\leq \| f\| \, \| g\| . \] The real-variable proof above is the standard direct proof of Cauchy–Schwarz specialized to this integral inner product.

The interval [𝑎, b] does not need to be finite. Bunyakovsky (Cauchy–Schwarz) holds on any measure space, including infinite intervals such as (-∞, ∞). What is essential is that the functions belong to 𝔏², i.e., their squares are integrable over the domain.

General form of the inequality: Let (X, 𝔐, μ) be any measure space. For any real (or complex) functions \( \displaystyle \quad f,g\in 𝔏^2(X), \quad \) the Bunyakovsky (Cauchy–Schwarz) inequality states: \[ \left| \int _Xf(x)\, g(x)\, d\mu (x)\right| \; \leq \; \left( \int _X|f(x)|^2\, d\mu (x)\right) ^{1/2}\left( \int _X|g(x)|^2\, d\mu (x)\right) ^{1/2}. \] This formulation does not require that X have finite measure.

Why finiteness of [𝑎, b] is irrelevant? The classical textbook statement uses [𝑎, b] simply because:

• it is the most familiar setting,
• continuity implies square-integrability on a compact interval.

But the proof itself uses only:

• the integral of a square is nonnegative,
• linearity of the integral,
• the discriminant argument for a quadratic polynomial.

None of these depend on the size of the interval. The only essential hypotheses are:

• f,g must be square integrable: \[ \int _X|f|^2<\infty ,\qquad \int _X|g|^2<\infty . \]
• The integral \( \displaystyle \quad \int_X fg \quad \) must be well-defined (which follows automatically from Hölder or from the inequality itself once f,g ∈ 𝔏²).

  Thus, the inequality holds on:

  • [𝑎, b] finite or infinite,
  • (0, ∞),
  • ℝⁿ,
  • any σ-finite measure space,
  • any Hilbert space with an inner product.

Since the Fourier transform G(ξ) of function g(x) belongs to 𝔏¹(ℝ), then g(x) is bounded as an inverse Fourier transform of G. The same property holds for its complex conjugate f^✶. Hence, its product f^✶·g is absolutely integrable. Let us evaluate its Fourier transform: \begin{align*} ℱ\left[ f^{\ast}\cdot g \right] &= \int_{-\infty}^{\infty} f^{\ast} (x)\,g(x)\, e^{-{\bf j}\xi x} {\text d}x \\ &= \int_{-\infty}^{\infty} f^{\ast} (x)\, e^{-{\bf j}\xi x} {\text d}x\,\frac{1}{2\pi} \int_{-\infty}^{\infty} G(\xi )\,e^{{\bf j}\xi x} {\text d}\xi \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\xi )\,{\text d}\xi \, \int_{-\infty}^{\infty} f^{\ast} (x)\, e^{-{\bf j}xi (x-u)} {\text d}x \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\xi )\,F^{\ast} (\xi - u)\.{\text d}\xi \end{align*} In the latter, we replaced f by ℱ⁻¹[F] and then used the property regarding complex conjugate of the integral can be transfered into its integrand. Finally, we set u = 0, which leads to the required property.

It is sufficient to set G(ξ) = F(ξ) in the previous Lemma.

It is clear that the second derivative f&prome;′ is bounded and belongs to 𝔏. Its Fourier Fourier is bounded on whole axis ℝ. Let us find its Fourier transform using integration by parts: \begin{align*} ℱ[f;;]_{x\to\xi} &= \int_{-\infty}^{\infty} f'' (x)\, e^{-{\bf j}\xi x} {\text d} x \\ &= \int_{-\infty}^{\infty} f' (x)\, e^{-{\bf j}\xi x} {\text d} x + \left. f' (x)\, e^{-{\bf j}\xi x} \right\vert_{x=-\infty}^{x=\infty} \\ &= \left( {\bf j}\xi \right)^2 \int_{-\infty}^{\infty} f (x)\, e^{-{\bf j}\xi x} {\text d} x = -\xi^2 \hat{f} (\xi ) . \end{align*} So we get an estimate: \[ \left\vert \xi^2 \hat{f} (\xi ) \right\vert = \left\vert ℱ[f''] \right\vert \le c \qquad\iff \qquad \left\vert \hat{f} (\xi ) \right\vert \frac{c}{|\xi |^2} . \] The latter inequality shows that ℱ[f] decreases at infinity as |ξ|⁻²; moreover, the Fourier transform of function f is a continuous function, so ℱ[f] ∈ 𝔏¹(ℝ).

Given the density of 𝒮𝒮(ℝ) in 𝔏²(ℝ). Setting u = v in Parseval’s formula, \[ \int_{-\infty}^{+\infty} \left( \hat{u}(\xi ) \right)^{\ast} \hat{v}(\xi )\,{\text d}\xi = \left( 2\pi \right) \int_{-\infty}^{+\infty} u^{\ast} (x)\,v(x)\,{\text d}x , \] gives