If $p$ and $q$ are nonnegative numbers such that $\frac{1}{p}+\frac{1}{q}=1$ and if $f \in L^p$ and $g \in L^q$, then $f\cdot g \in L^1$ and $$\int |fg| \leqslant ||f||_p \cdot ||g||_q$$

I think Hölder's inequality is derived in order to prove Minkowski's inequality, which is a generalization of triangle inequality to $L^p$ norm. But is there any intuitive understanding of Hölder's inequality? It's hard for me to remember it. It seems that it's a generalization of the Cauchy-Schwarz inequality, trying to compare $L^2$ inner product to norm, but the power of each term is different, which makes it harder to be understood compared with Minkowski inequality.


2 Answers 2


Hölder's inequality is an attempt to generalize the Cauchy-Schwarz inequality to other Lebesgue exponents (other $L^p$ norm). In fact, for inequality of the form $$ \|fg\|_1 \leq \|f\|_p \|g\|_q$$ to be true, we must have $1/p+1/q=1$. To see this, we use scaling argument, which is also useful to verify the correct exponent for other type of inequalities (Sobolev inequalities, etc.).

If such inequality were true, then we could apply the inequality to functions $f(\lambda x)$ and $g(\lambda x)$ for $\lambda\in \mathbb{R}$ to get $$ \|fg\|_1 \leq \lambda^{n(1-1/p-1/q)} \|f\|_p \|g\|_q$$ which cannot be true for all $\lambda$ unless $1/p+1/q=1$. This is a way to see that indeed the correct exponents must be such that $1/p+1/q=1$.

  • $\begingroup$ Sorry, but can you explain why applying to $f(\lambda x)$ and $g(\lambda x)$ gives $ \|fg\|_1 \leq \lambda^{n(1-1/p-1/q)} \|f\|_p \|g\|_q$? how did you take the $\lambda$ out of the norms? $\endgroup$
    – lady gaga
    Apr 29, 2021 at 13:25
  • 1
    $\begingroup$ @900edges I believe he is thinking of functions $f, g$ on $\mathbb{R}^n$, and applying the change of variables $y = \lambda x$ (which has Jacobian $\lambda^n$, so $dx = \lambda^{-n} dy$ and then you can easily factor the relevant power of $\lambda$ out of the integral -- in the $p, q$ norms you get a $p$ or $q$ root due to their definitions). $\endgroup$
    – Pedro
    May 4, 2021 at 2:00
  • $\begingroup$ @Pedro and digiboy1 I believe that this Jacobian-related argument ought to be fleshed out in order for this to become a bit more convincing. At the moment we don't even have any idea where the $n$ came from, in the main answer $\endgroup$ Nov 22, 2021 at 14:45

Slightly late to the game, but just in case this is helpful:

For sequences, Cauchy-Schwarz asserts that $$ (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2+b_2^2+ \cdots + b_n^2) $$ Or in words: $$ \textit{the square of the sum of products} \leq \textit{the product of the sum of squares} $$ Staring at this mnemonic, one can imagine many ways it might generalize. For example, maybe the cube of the sum of products is bounded by the product of the sum of cubes? Turns out this is true, with two small caveats:

  • Cauchy-Schwarz involves squares and products of two sequences. To generalize to cubes, we need three sequences.
  • We need to require all our sequences to be non-negative.

In symbols, the cubic generalization of Cauchy-Schwarz reads: $$ (a_1 b_1 c_1 + a_2 b_2 c_2 + \cdots + a_n b_n c_n)^3 \leq (a_1^3 + a_2^3 + \cdots + a_n^3)(b_1^3+b_2^3+ \cdots + b_n^3) (c_1^3 + c_2^3 + \cdots + c_n^3) $$ Hölder's inequality is the natural extension of this idea: for any positive integer $n$, the $n$-th power of the sum of products (of $n$ non-negative sequences) is bounded by the product of the sum of the $n$-th powers. It's a fun exercise to deduce the usual statement of Hölder's inequality for $p,q \in \mathbb{Q}$ from this one. This in turn forces Hölder's inequality to hold for all positive $p,q \in \mathbb R$.

  • $\begingroup$ It is not obvious how your consideration of three vectors relates to the statement of Holder's inequality (in Euclidean spaces) which involves two vectors and not three $\endgroup$ Nov 22, 2021 at 14:41

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