Using convexity in the proof of Hölder’s inequality
A key fact for the algebra properties of $L^p$ spaces is Hölder’s inequality:
Let $f \in L^p$ and $g \in L^q$ for some $0 < p,q \leq \infty$. Then $fg \in L^r$ and $\|fg\|_{L^r} \leq \|f\|_{L^p} \|g\|_{L^q}$, where the exponent $r$ is defined by the formula $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$.
The endpoint cases where $p=\infty$ or $q=\infty$ can be proved separately and easily, usually left as exercises for dealing with the $L^\infty$ norm in textbooks for real analysis.
There are many known proofs for this theorem, including one from the Wikipedia link above.
This post focuses on the specific proof in Terry Tao's real analysis lecture notes.
In his notes, Terry Tao reduces the statement, using homogeneity, to the nontrivial essential cases where $p,q<\infty$, $r=1$, and $\|f\|_{L^p}=\|g\|_{L^q} = 1$. I very much like how he isolates the essential component of the proof by reducing the statement to simpler cases, which is conceptually (and pedagogically) natural to understand and is a problem technique used often.
The following is his proof:
Our task is now to show that $$\int_X |fg|\ d\mu \leq 1. \tag{1}$$ Here, we use the convexity of the exponential function $t \mapsto e^t$ on ${}[0,+\infty)$, which implies the convexity of the function $t \mapsto |f(x)|^{p(1-t)} |g(x)|^{qt}$ for $t \in [0,1]$ for any $x$. In particular, we have $$ |f(x) g(x)| \leq \frac{1}{p} |f(x)|^p + \frac{1}{q} |g(x)|^q \tag{2} $$ and the claim $(1)$ follows from the normalisations on $p, q, f, g.$
Questions:
- How does the convexity of the exponential function imply the convexity of the function $t \mapsto |f(x)|^{p(1-t)} |g(x)|^{qt}$ for $t \in [0,1]$ for any $x$?
- How does one get (2)?
Notes. Answers to these questions are rather trivial to experienced readers but may require some effort for beginners. I will write my own answers below.
1 answer
The first question can be rephrased in a clear way as follows:
Let $A$ and $B$ be two positive real numbers. Show that the function $F(t):=A^{p(1-t)}B^{qt}$ for $t\in[0,1]$ is convex using the convexity of the exponential function.
If one rewrites: $$ A^{p(1-t)}B^{qt} = e^{p(1-t)\log(A)+(qt)\log(B)} $$ where the exponent is an affine function $h(t)= at+b$, and one can easily check that $$ h(\lambda t_1+(1-\lambda)t_2) = \lambda h(t_1)+(1-\lambda)h(t_2) $$ Now, one can easily follow the convexity of the exponential function to show that $F(t)=e^{h(t)}$ is convex.
To see (2) as a particular case from convexity of $F$, one only needs to write $$ F(\lambda t_1+(1-\lambda)t_2)\le\lambda F(t_1)+(1-\lambda)F(t_2) $$ with $$ \lambda = \frac1p, t_1 = 0, t_2=1\ . $$
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