The GOATs
\begin{align*} {\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert} = \leq {\left\lVert {f} \right\rVert}_{2} {\left\lVert {g} \right\rVert}_{2} && \text{with equality} \iff f = \lambda g .\end{align*}
In general, Cauchy-Schwarz relates inner product to norm, and only happens to relate norms in \(L^1\). Some other useful forms: \begin{align*} \left(\sum_{k=1}^{n} a_{k} b_{k}\right)^{2} &\leq\left(\sum_{k=1}^{n} a_{k}^{2}\right)\left(\sum_{k=1}^{n} b_{k}^{2}\right) \\ \left|\int_{\mathbb{R}^{n}} f(x) \overline{g(x)} d x\right|^{2} &\leq \int_{\mathbb{R}^{n}}|f(x)|^{2} d x \int_{\mathbb{R}^{n}}|g(x)|^{2} d x .\end{align*}
\begin{align*} {\left\lvert {{\left\lVert {x} \right\rVert} - {\left\lVert {y} \right\rVert}} \right\rvert} \leq {\left\lVert {x - y} \right\rVert} .\end{align*}
\begin{align*} \frac 1 p + \frac 1 q = 1 \implies {\left\lVert {f g} \right\rVert}_{1} \leq {\left\lVert {f} \right\rVert}_{p} {\left\lVert {g} \right\rVert}_{q} .\end{align*}
With integrals: \begin{align*} \int_X {\left\lvert {fg} \right\rvert} \leq \qty{\int_X {\left\lvert {f} \right\rvert}^p}^{1\over p} \qty{\int_X {\left\lvert {f} \right\rvert}^q}^{1\over q} .\end{align*}
It suffices to show this when \({\left\lVert {f} \right\rVert}_p = {\left\lVert {g} \right\rVert}_q = 1\), since \begin{align*} \|f g\|_{1} \leq\|f\|_{p}\|f\|_{q} \Longleftrightarrow \int \frac{|f|}{\|f\|_{p}} \frac{|g|}{\|g\|_{q}} \leq 1 .\end{align*}
Using \(AB \leq \frac 1 p A^p + \frac 1 q B^q\), we have \begin{align*} \int|f \| g| \leq \int \frac{|f|^{p}}{p} \frac{|g|^{q}}{q}=\frac{1}{p}+\frac{1}{q}=1 .\end{align*}
For finite measure spaces, \begin{align*} 1 \leq p < q \leq \infty \implies L^q \subset L^p \quad (\text{ and } \ell^p \subset \ell^q) .\end{align*}
Fix \(p, q\), let \(r = \frac q p\) and \(s = \frac{r}{r-1}\) so \(r^{-1}+ s^{-1}= 1\). Then let \(h = {\left\lvert {f} \right\rvert}^p\):
\begin{align*} {\left\lVert {f} \right\rVert}_{p}^p = {\left\lVert {h\cdot 1} \right\rVert}_{1} \leq {\left\lVert {1} \right\rVert}_{s} {\left\lVert {h} \right\rVert}_{r} = \mu(X)^{\frac 1 s} {\left\lVert {f} \right\rVert}_{q}^{\frac q r} \implies {\left\lVert {f} \right\rVert}_{p} \leq \mu(X)^{\frac 1 p - \frac 1 q} {\left\lVert {f} \right\rVert}_{q} .\end{align*}
Note: doesn’t work for \(\ell_p\) spaces, but just note that \(\sum {\left\lvert {x_n} \right\rvert} < \infty \implies x_n < 1\) for large enough \(n\), and thus \(p<q \implies {\left\lvert {x_n} \right\rvert}^q \leq {\left\lvert {x_n} \right\rvert}^q\).
For \(x\in H\) a Hilbert space and \(\left\{{e_k}\right\}\) an orthonormal sequence, \begin{align*} \sum_{k=1}^{\infty}\| {\left\langle {x},~{e_{k} } \right\rangle} \|^{2} \leq \|x\|^{2} .\end{align*}
Note that this does not need to be a basis.
Equality in Bessel’s inequality, attained when \(\left\{{e_k}\right\}\) is a basis, i.e. it is complete, i.e. the span of its closure is all of \(H\). This states that if \(\left\{{e_k}\right\}\) is an orthonormal basis for \(H\), then \begin{align*} \sum_{k\geq 0} {\left\lvert { {\left\langle {x},~{e_k} \right\rangle} } \right\rvert} ^2 = {\left\lVert {x} \right\rVert}_H^2 .\end{align*}
This appears in several other forms: \begin{align*} {1\over 2\pi} \int_{(-\pi, \pi)} {\left\lvert {f} \right\rvert}^2 = \sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert}^2 && c_k \coloneqq{1\over 2\pi } \int_{(-\pi, \pi)} f(x) e^{-ikx} \,dx .\end{align*}
\begin{align*} {\left\lVert {f} \right\rVert}_{L^2}^2 &= {\left\lVert {\widehat{f}} \right\rVert}_{L^2} \\ \int_{{\mathbf{R}}^d} {\left\lvert {f} \right\rvert}^2 &= \int_{{\mathbf{R}}^d} {\left\lvert {\widehat{f}} \right\rvert}^2 .\end{align*}
Slogan: the \(L^2\) norm of the function equals the \(\ell^2\) norm of the Fourier coefficients
Less common
The most often used form here: \begin{align*} \mu \qty{ f^{-1}\qty{(\alpha, \infty)} } \coloneqq\mu\qty{\left\{{ x\in X {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} > \alpha }\right\}} \leq {1\over \alpha} {\left\lVert {f} \right\rVert}_1 \coloneqq{1\over \alpha} \int_X {\left\lvert {f} \right\rvert} .\end{align*} Proof: let \(S_\alpha\) be the set appearing, then \(\alpha \mu(S_\alpha)\) is the sum of areas of certain boxes below the graph of \(f\). Interpret \(\int_X f\) as the total area under the graph to make the inequality obvious.
The probability interpretation: \({\mathbf{P}}(X\geq \alpha) \leq {1\over \alpha} {\mathbb{E}}(X)\).
The more general version: \begin{align*} \mu \qty{ f^{-1}\qty{(\alpha, \infty)} } \coloneqq\mu\qty{\left\{{ x\in X {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} > \alpha }\right\} } \leq {1\over \alpha^p} {\left\lVert {f} \right\rVert}_p^p \coloneqq{1\over \alpha^p} \int_X {\left\lvert {f} \right\rvert}^p .\end{align*} Proof: \begin{align*} {\left\lVert {f} \right\rVert}_p^p = \int {\left\lvert {f} \right\rvert}^p \geq \int_{S_\alpha} {\left\lvert {f} \right\rvert}^p \geq \alpha^p \int_{S_\alpha} 1 = \alpha^p \mu(S_\alpha) .\end{align*}
\begin{align*} 1\leq p < \infty \implies {\left\lVert {f+g} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}_{p}+ {\left\lVert {g} \right\rVert}_{p} .\end{align*}
This does not handle \(p=\infty\) case. Use to prove \(L^p\) is a normed space.
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We first note \begin{align*} {\left\lvert {f+g} \right\rvert}^p = {\left\lvert {f+g} \right\rvert}{\left\lvert {f+g} \right\rvert}^{p-1} \leq \left( {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\right) {\left\lvert {f+g} \right\rvert}^{p-1} .\end{align*}
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Note that if \(p,q\) are conjugate exponents then \begin{align*} \frac 1 q &= 1 - \frac 1 p = \frac{p-1} p \\ q &= \frac p {p-1} .\end{align*}
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Then taking integrals yields \begin{align*} {\left\lVert {f+g} \right\rVert}_p^p &= \int {\left\lvert {f+g} \right\rvert}^p \\ &\leq \int \left( {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\right) {\left\lvert {f+g} \right\rvert}^{p-1} \\ &= \int {\left\lvert {f} \right\rvert} {\left\lvert {f+g} \right\rvert}^{p-1} + \int {\left\lvert {g} \right\rvert} {\left\lvert {f+g} \right\rvert}^{p-1} \\ &= {\left\lVert {f(f+g)^{p-1}} \right\rVert}_1 + {\left\lVert {g(f+g)^{p-1}} \right\rVert}_1 \\ &\leq {\left\lVert {f} \right\rVert}_p ~{\left\lVert {(f+g)^{p-1})} \right\rVert}_q + {\left\lVert {g} \right\rVert}_p ~{\left\lVert {(f+g)^{p-1})} \right\rVert}_q \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) {\left\lVert { (f+g)^{p-1})} \right\rVert}_q \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \left( \int {\left\lvert {f+g} \right\rvert}^{(p-1)q} \right)^{\frac 1 q} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \left( \int {\left\lvert {f+g} \right\rvert}^{p} \right)^{1 - \frac 1 p} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{\int {\left\lvert {f+g} \right\rvert}^{p} }{\left( \int {\left\lvert {f+g} \right\rvert}^{p} \right)^{\frac 1 p}} \\ &= \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{{\left\lVert {f+g} \right\rVert}_p^p}{{\left\lVert {f+g} \right\rVert}_p} .\end{align*}
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Cancelling common terms yields \begin{align*} 1 &\leq \left( {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p \right) \frac{1}{{\left\lVert {f+g} \right\rVert}_p} \\ &\implies {\left\lVert {f+g} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p .\end{align*}
\begin{align*} \frac 1 p + \frac 1 q = \frac 1 r + 1 \implies \|f \ast g\|_{r} \leq\|f\|_{p}\|g\|_{q} \end{align*}
\begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 & \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1 \\ \|f * g\|_{p} & \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}p, \\ {\left\lVert {f\ast g} \right\rVert}_\infty & \leq {\left\lVert {f} \right\rVert}_2 {\left\lVert {g} \right\rVert}_2 \\ {\left\lVert {f\ast g} \right\rVert}_\infty & \leq {\left\lVert {f} \right\rVert}_p {\left\lVert {g} \right\rVert}_q .\end{align*}
Inequalities that appear in proofs
\begin{align*} \sqrt{ab} \leq \frac{a+b}{2} .\end{align*}
\begin{align*} f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) .\end{align*}
\begin{align*} AB \leq {A^p \over p} + {B^q \over q} .\end{align*}
\begin{align*} (a+b)^p \leq 2^{p-1} (a^p + b^p) .\end{align*}
\begin{align*} (1 + x)^n \geq 1 +nx \quad x\geq -1, \text{ or } n\in 2{\mathbf{Z}}\text{ and } \forall x .\end{align*}
As a consequence, \begin{align*} 1-x \leq e^{-x} .\end{align*}
\begin{align*} \forall t\in {\mathbf{R}},\quad 1 + t \leq e^t .\end{align*}
- It’s an equality when \(t=0\).
- \({\frac{\partial }{\partial t}\,} 1+ t < {\frac{\partial t}{\partial e}\,}^t \iff t<0\)
\begin{align*} {1\over r} \coloneqq{1\over p} + {1\over q} - 1 \implies {\left\lVert {f \ast g} \right\rVert}_{r} \leq {\left\lVert {f} \right\rVert}_{p} {\left\lVert {g} \right\rVert}{q} .\end{align*}
- \(\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}\).