11. Maximum principle in unbounded regions and the Phragmen–Lindelöf method PDF TEX
Phragmen–Lindelöf method
Maximum modulus principle
Let \(f\) be defined on \(\Omega\) and \(a \in \Omega\). Then \(|f|\) has a local maximum at \(a\) if there exists \(\delta>0\) such that \(D(a, \delta) \subseteq \Omega\) and \(|f(a)| \geq|f(z)|\) for every \(z \in D(a, \delta)\). Further, we say that \(|f|\) has no local maximum in \(\Omega\) if \(|f|\) does not have local maximum at every point of \(\Omega\). Similarly, we define a local minimum.
Suppose that \(\Omega\) is a region and \(f \in H(\Omega)\).
Then \(|f|\) has no local maximum at any point of \(\Omega\), unless \(f\) is constant.
Moreover, if the closure of \(\Omega\) is compact and \(f\) is continuous on \(\overline{\Omega}\), then \[\sup_{z\in\Omega}|f(z)|\le \sup_{z\in\partial\Omega}|f(z)|.\]
The boundedness of \(\Omega\) in the previous theorem is necessary
Remark.
Thus the maximum modulus principle is not valid in unbounded open sets. We elaborate it by the following example. Let \[\Omega=\left\{z=x+i y: -\frac{\pi}{2}<y<\frac{\pi}{2}\right\}\] and \[f(z)=\exp (\exp (z)).\]
Then \(\Omega\) is unbounded and \(\max _{z \in \partial \Omega}|f(z)|=1\), since \[\left|f\left(x \pm \frac{\pi}{2} i\right)\right|=\left|\exp \left(\exp \left(x \pm \frac{\pi}{2} i\right)\right)\right|=|\exp ( \pm i \exp (x))|=1.\]
On the other hand, \(f(x)=e^{e^{x}} \rightarrow \infty\) as \(x\) tends to infinity through positive reals.
Phragmen–Lindelöf method
It has already been pointed out that the maximum modulus principle need not be valid in unbounded regions.
We will illustrate the Phragmen–Lindelöf method, which will enable to us to prove that a function is constant in a region whenever we control its growth in the region.
We need not necessarily assume that a function is bounded in the region for concluding that it is constant as is the case of the Liouville theorem.
For example, an entire function \(f(z)\) is constant if \[|f(z)| \leq 1+|z|^{\frac{1}{2}}\] for \(z \in \mathbb{C}\).
For given \(a, b \in \mathbb{R}\), let \[\Omega=\{(x+i y) : a<x<b\}.\]
Let \(f\) be continuous on \(\overline{\Omega}\) and \(f \in H(\Omega)\) and assume that \(|f(z)|<B\) for \(z \in \Omega\) and fixed \(B>0\). For a \(\leq x \leq b\), let \[M(x)=\sup \{|f(x+i y)|: -\infty<y<\infty\}.\] Then we have \[(M(x))^{b-a} \leq(M(a))^{b-x}(M(b))^{x-a} \quad \text { for } \quad a \leq x \leq b. \tag{*}\]
Proof: For \(\varepsilon>0\), we consider \(f+\varepsilon\) in place of \(f\) if \(M(a)=0\) and \(f-\varepsilon\) in place of \(f\) if \(M(b)=0\) and let \(\varepsilon\) tend to zero to observe that there is no loss of generality in assuming that \(M(a)>0\) and \(M(b)>0\).
For \(z \in \overline{\Omega}\), we write \(z=x+i y\) with \(a \leq x \leq b\).
Claim Suppose that the theorem is valid for all functions \(f\) satisfying the assumptions of the theorem together with \(M(a)=M(b)=1\).
We now prove the general case. Let \[g(z)=(M(a))^{\frac{b-z}{b-a}}(M(b))^{\frac{z-a}{b-a}}.\]
We observe that \(g(z)\) is analytic in \(\mathbb{C}\) and it has no zero in \(\mathbb{C}\). Further \[|g(z)|=(M(a))^{\frac{b-x}{b-a}}(M(b))^{\frac{x-a}{b-a}},\] and the exponents on the right-hand side above lie in \([0,1]\).
Therefore \[|g(z)| \geq \tau \quad \text { for } \quad z \in \overline{\Omega}\] where \[\tau=\min (1, M(a)) \min (1, M(b))>0,\] and \[|g(a+i y)|=M(a),\quad \text{ and } \quad |g(b+i y)|=M(b).\]
Now we consider \(h(z)=\frac{f(z)}{g(z)}\). By the definition of \(h\) and our assumption \(M(a)= M(b)=1\), we see that \[\sup _{-\infty<y<\infty}|h(a+i y)|=\sup _{-\infty<y<\infty}|h(b+i y)|=1,\] and \[|h(z)| \leq B \tau^{-1} \quad \text { for }\quad z \in \Omega.\]
Therefore \(h\) satisfies the assumptions of the theorem. Hence \[|h(x+i y)| \leq 1 \quad \text { for }\quad a \leq x \leq b.\]
Using \(|g(z)|=(M(a))^{\frac{b-x}{b-a}}(M(b))^{\frac{x-a}{b-a}}\) and \(|h(x+i y)| \leq 1\), we have \[|f(x+i y)| \leq|g(x+i y)|=(M(a))^{\frac{b-x}{b-a}}(M(b))^{\frac{x-a}{b-a}}\quad \text { for } \quad a \leq x \leq b,\] which implies (*).
It remains to prove the theorem when \(M(a)=M(b)=1\), which we assume from now on and we complete the proof by showing that \(M(x) \leq 1\) for \(a \leq x \leq b\).
Since \(f\) is continuous on \(\overline{\Omega}\) and \(|f(z)|<B\) for \(z \in \Omega\), we see that \[|f(z)| \leq B\quad \text { for }\quad z \in \overline{\Omega}.\]
For \(\varepsilon>0\), we define \[h_{\varepsilon}(z)=\frac{1}{1+\varepsilon(z-a)}.\]
For a fixed \(z\in\mathbb C\), we see that \(\lim_{\varepsilon \to 0}h_{\varepsilon}(z) = 1\).
Therefore it suffices to prove for every \(\varepsilon>0\) we have \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1 \quad \text { for }\quad z \in \overline{\Omega}.\]
First, we estimate \(h_{\varepsilon}(z)\). We observe that \[\operatorname{Re}(1+\varepsilon(z-a)) \geq 1 \quad \text { for } \quad z \in \overline{\Omega}.\]
Therefore \[\left|h_{\varepsilon}(z)\right| \leq \frac{1}{\operatorname{Re}(1+\varepsilon(z-a))} \leq 1 \quad \text { for } \quad z \in \overline{\Omega}.\tag{A}\]
Next \[\begin{aligned} |\operatorname{Im}(1+\varepsilon(z-a))| \geq \varepsilon|y|\quad \text { for } \quad z \in \overline{\Omega}, \end{aligned}\] and therefore \[\begin{aligned} \left|f(z) h_{\varepsilon}(z)\right| \leq \frac{B}{\varepsilon|y|} \quad \text { for } \quad z \in \overline{\Omega}. \end{aligned}\tag{B}\]
In particular \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1\quad \text { if } \quad z \in \overline{\Omega}\quad \text { and }\quad |\operatorname{Im}(z)| \geq \frac{B}{\varepsilon}.\]
We consider the closed set \(S\) consisting of rectangle as here
Since \(M(a)=M(b)=1\), we see from (A) and (B) that \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1 \text { for } z \in \partial S\]
Therefore by the maximum principle, we have \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1 \text { for } z \in S.\]
Also by (B) we conclude that \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1 \text { for } z \in \overline{\Omega} \backslash S.\]
Hence \[\left|f(z) h_{\varepsilon}(z)\right| \leq 1 \text { for } z \in \overline{\Omega}.\]
This proves the theorem.$$\tag*{$\blacksquare$}$$
Suppose that the assumptions of the previous theorem are satisfied. Further suppose that \(f\) is not constant. Then \[|f(z)|<\max (M(a), M(b)) \quad \text { for } \quad z \in \Omega.\]
Proof: Assume that \(M(a) \neq M(b)\). By the previous theorem, we have \[M(x)^{b-a}<(\max (M(a), M(b)))^{b-a} \quad \text{ for } \quad a \leq x \leq b.\] Therefore, the assertion follows, since \[M(x)<\max (M(a), M(b)) \quad \text{ for } \quad a \leq x \leq b.\] Let \(M(a)=M(b)\). Then, by the previous theorem as above, we have \(M(x) \leq M(a)\) for \(a \leq x \leq b\). We fix arbitrary \(z_{0} \in \Omega\). Then there is \(r>0\) such that \(\overline{D}\left(z_{0}, r\right) \subset \Omega\). By the maximum principle, we have \[\left|f\left(z_{0}\right)\right|<\max _{\left|z-z_{0}\right|=r}|f(z)| \leq M(x) \leq M(a)\quad \text{ for some } \quad a < x < b. \quad\tag*{$\blacksquare$}\]
Hadamard three-circle theorem
Let \(0<R_{1}<R_{2}<\infty\) and \[A\left(0 ; R_{1}, R_{2}\right)=\left\{z\in\mathbb C: R_{1}<|z|<R_{2}\right\}.\] Let \(g(z)\) be continuous in \(\overline{A}\left(0 ; R_{1}, R_{2}\right)\) and holomorphic in \(A\left(0 ; R_{1}, R_{2}\right)\). Let \[B(R)=\max _{|z|=R}|g(z)| \quad \text { for } \quad R_{1} \leq R \leq R_{2} .\]
Then \[(B(R))^{\log R_{2}-\log R_{1}} \leq\left(B\left(R_{1}\right)\right)^{\log R_{2}-\log R}\left(B\left(R_{2}\right)\right)^{\log R-\log R_{1}}\] for \(R_{1} \leq R \leq R_{2}\).
Proof: Let \(\Omega=\left\{x+i y\in\mathbb C: \log R_{1}<x<\log R_{2}\right\}\) and let \(e(z)=e^{z}\).
We observe that \(e(z) \in \overline{A}\left(0 ; R_{1}, R_{2}\right)\) for \(z \in \overline{\Omega}\). Thus \(e\) is a function from \(\overline{\Omega}\) into \(\overline{A}\left(0 ; R_{1}, R_{2}\right)\). Further it is onto.
Moreover it maps a vertical line in \(\overline{\Omega}\) passing through \((x, 0)\) onto a circle \(\left\{z\in \mathbb C: |z|=e^{x}\right\}\) in \(\overline{A}\left(0 ; R_{1}, R_{2}\right)\).
Next we write \(f(z)=g(e(z))\) for \(z \in \overline{\Omega}\).
We observe that \(f\) is continuous in \(\overline{\Omega}\) and analytic in \(\Omega\). Since \(\overline{A}\left(0 ; R_{1}, R_{2}\right)\) is compact, we see that \(f\) is bounded on \(\overline{\Omega}\).
Thus \(f\) satisfies all the assumptions of the previous theorem. Further for \(R_{1} \leq R \leq R_{2}\), we have \[\begin{aligned} M(\log R) & =\sup \{|f(\log R+i y)|: -\infty<y<\infty\} \\ & =\sup \left\{g\left(R e^{i \theta}\right) : 0 \leq \theta \leq 2 \pi\right\}=B(R). \end{aligned}\]
Hence we conclude from the previous theorem with \(a=\log R_{1}\) and \(b=\log R_{2}\) that \[(M(\log R))^{\log R_{2}-\log R_{1}} \leq\left(M\left(\log R_{1}\right)\right)^{\log R_{2}-\log R}\left(M\left(\log R_{2}\right)\right)^{\log R-\log R_{1}}.\]
This together with \[\begin{aligned} M(\log R) & =\sup \{|f(\log R+i y)|: -\infty<y<\infty\} \\ & =\sup \left\{g\left(R e^{i \theta}\right) : 0 \leq \theta \leq 2 \pi\right\}=B(R), \end{aligned}\] implies that \[(B(R))^{\log R_{2}-\log R_{1}} \leq\left(B\left(R_{1}\right)\right)^{\log R_{2}-\log R}\left(B\left(R_{2}\right)\right)^{\log R-\log R_{1}}\] for \(R_{1} \leq R \leq R_{2}\).
This completes the proof of Hadamard’s theorem. $$\tag*{$\blacksquare$}$$
Phragmen–Lindelöf method: applications
Suppose \[\Omega=\left\{x+i y\in\mathbb C:|y|<\frac{\pi}{2}\right\}.\] Suppose \(f\) is continuous on \(\overline{\Omega}\), and \(f \in H(\Omega)\), and there are constants \(\alpha<1\), \(A<\infty\), such that \[|f(z)|<\exp \{A \exp (\alpha|x|)\}, \quad\text{ whenever } \quad z=x+i y \in \Omega,\] and \[\left|f\left(x \pm \frac{\pi i}{2}\right)\right| \leq 1 \quad \text{ for }\quad -\infty<x<\infty.\] Then \(|f(z)| \leq 1\) for all \(z \in \Omega\).
Note that the conclusion does not follow if \(\alpha=1\), as is shown by the function \(\exp (\exp z)\).
Proof: Choose \(\beta>0\) so that \(\alpha<\beta<1\). For \(\varepsilon>0\), define \[h_{\varepsilon}(z)=\exp \left\{-\varepsilon\left(e^{\beta z}+e^{-\beta z}\right)\right\}.\]
For \(z \in \overline{\Omega}\), we have
\[\operatorname{Re}\left[e^{\beta z}+e^{-\beta z}\right]=\left(e^{\beta x}+e^{-\beta x}\right) \cos \beta y \geq \delta\left(e^{\beta x}+e^{-\beta x}\right),\] where \(\delta=\cos (\beta \pi / 2)>0\), since \(|\beta|<1\). Hence \[\left|h_{\varepsilon}(z)\right| \leq \exp \left\{-\varepsilon \delta\left(e^{\beta x}+e^{-\beta x}\right)\right\}<1 \quad \text{ for } \quad z \in \overline{\Omega}.\]
It follows that \(\left|f h_{\varepsilon}\right| \leq 1\) on \(\partial \Omega\) and that
\[\left|f(z) h_{\varepsilon}(z)\right| \leq \exp \left\{A e^{\alpha|x|}-\varepsilon \delta\left(e^{\beta x}+e^{-\beta x}\right)\right\} \quad \text{ for } \quad z \in \overline{\Omega}.\]
Fix \(\varepsilon>0\). Since \(\varepsilon \delta>0\) and \(\beta>\alpha\), we have \[\lim_{x\to\pm\infty}\Big(A e^{\alpha|x|}-\varepsilon \delta\left(e^{\beta x}+e^{-\beta x}\right)\Big)=-\infty\]
Hence there exists an \(x_{0}\) so that \[\left|f(z) h_{\varepsilon}(z)\right|\le 1 \quad \text{ for all } \quad x>x_0.\]
Since \(\left|f h_{\varepsilon}\right| \leq 1\) on the boundary of the rectangle whose vertices are \(\pm x_{0} \pm(\pi i / 2)\), the maximum modulus theorem shows that actually \(\left|f h_{\varepsilon}\right| \leq 1\) on this rectangle.
Thus \(\left|f h_{\varepsilon}\right| \leq 1\) at every point of \(\Omega\), for every \(\varepsilon>0\). \[\lim_{\varepsilon \to 0}h_{\varepsilon}(z)=1 \quad \text{ for every }\quad z\in\mathbb C,\] so we conclude that \(|f(z)| \leq 1\) for all \(z \in \Omega\) as desired. $$\tag*{$\blacksquare$}$$
Applications in harmonic analysis
Riesz interpolation theorem
Let \((X, \mu)\) and \((Y, v)\) be two \(\sigma\)-finite measure spaces. Let \(T\) be a linear operator defined on the set of all finitely simple functions on \(X\) and taking values in the set of measurable functions on \(Y\). Let \(1 \leq p_{0}, p_{1}, q_{0}, q_{1} \leq \infty\) and assume that \[\begin{aligned} \|T(f)\|_{L^{q_{0}}} \leq M_{0}\|f\|_{L^{p_{0}}}, \quad \text{ and } \quad \|T(f)\|_{L^{q_{1}}} \leq M_{1}\|f\|_{L^{p_{1}}}, \end{aligned}\] for all finitely simple functions \(f\) on \(X\). Then for all \(0<\theta<1\) we have \[\|T(f)\|_{L^{q}} \leq M_{0}^{1-\theta} M_{1}^{\theta}\|f\|_{L^{p}} \tag{*}\] for all finitely simple functions \(f\) on \(X\), where \[\frac{1}{p}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}} \quad \text { and } \quad \frac{1}{q}=\frac{1-\theta}{q_{0}}+\frac{\theta}{q_{1}} . \tag{**}\]
Remarks.
Recall that a simple function is called finitely simple if it is supported in a set of finite measure. Finitely simple functions are dense in \(L^{p}(X, \mu)\) for \(0<p<\infty\), whenever \((X, \mu)\) is a \(\sigma\)-finite measure space.
Consequently, when \(p<\infty\), by density, \(T\) has a unique bounded extension from \(L^{p}(X, \mu)\) to \(L^{q}(Y, v)\) when \(p\) and \(q\) are as in (**).
The proof will be based on the following lemma:
Let \(F\) be analytic in the open strip \(S=\{z \in \mathbb{C}: 0<\operatorname{Re} z<1\}\), continuous and bounded on its closure, such that \(|F(z)| \leq B_{0}\) when \(\operatorname{Re} z=0\) and \(|F(z)| \leq B_{1}\) when \(\operatorname{Re} z=1\), for some \(0<B_{0}, B_{1}<\infty\). Then \(|F(z)| \leq B_{0}^{1-\theta} B_{1}^{\theta}\) when \(\operatorname{Re} z=\theta\), for any \(0 \leq \theta \leq 1\).
Proof: Exercise! $$\tag*{$\blacksquare$}$$
Proof: Let \[f=\sum_{k=1}^{m} a_{k} e^{i \alpha_{k}} \chi_{A_{k}}\] be a finitely simple function on \(X\), where \(a_{k}>0, \alpha_{k}\) are real, and \(A_{k}\) are pairwise disjoint subsets of \(X\) with finite measure.
We need to control \[\|T(f)\|_{L^{q}(Y, v)}=\sup _{g}\left|\int_{Y} T(f)(y) g(y) d v(y)\right|,\] where the supremum is taken over all finitely simple functions \(g\) on \(Y\) with \(L^{q^{\prime}}\) norm less than or equal to \(1\).
Write \[g=\sum_{j=1}^{n} b_{j} e^{i \beta_{j}} \chi_{B_{j}},\] where \(b_{j}>0, \beta_{j}\) are real, and \(B_{j}\) are pairwise disjoint subsets of \(Y\) with finite \(v\) measure.
Let \[P(z)=\frac{p}{p_{0}}(1-z)+\frac{p}{p_{1}} z \quad \text { and } \quad Q(z)=\frac{q^{\prime}}{q_{0}^{\prime}}(1-z)+\frac{q^{\prime}}{q_{1}^{\prime}} z.\]
For \(z\) in the closed strip \(\overline{S}=\{z \in \mathbb{C}: 0 \leq \operatorname{Re} z \leq 1\}\), define \[f_{z}=\sum_{k=1}^{m} a_{k}^{P(z)} e^{i \alpha_{k}} \chi_{A_{k}}, \quad \text{ and }\quad g_{z}=\sum_{j=1}^{n} b_{j}^{Q(z)} e^{i \beta_{j}} \chi_{B_{j}},\] and \[F(z)=\int_{Y} T\left(f_{z}\right)(y) g_{z}(y) d v(y)\]
Notice that \(f_{\theta}=f\) and \(g_{\theta}=f\). By linearity we have \[F(z)=\sum_{k=1}^{m} \sum_{j=1}^{n} a_{k}^{P(z)} b_{j}^{Q(z)} e^{i \alpha_{k}} e^{i \beta_{j}} \int_{Y} T\left(\chi_{A_{k}}\right)(y) \chi_{B_{j}}(y) d v(y)\]
Since \(a_{k}, b_{j}>0, F\) is analytic in \(z\), and the expression \[\int_{Y} T\left(\chi_{A_{k}}\right)(y) \chi_{B_{j}}(y) d v(y)\] is a finite constant, being an absolutely convergent integral; this is seen by Hölder’s inequality with exponents \(q_{0}\) and \(q_{0}^{\prime}\) (or \(q_{1}\) and \(q_{1}^{\prime}\) ) and our assumption \[\|T(f)\|_{L^{q_{0}}} \leq M_{0}\|f\|_{L^{p_{0}}}, \quad \text{ and } \quad \|T(f)\|_{L^{q_{1}}} \leq M_{1}\|f\|_{L^{p_{1}}}.\]
By the disjointness of the sets \(A_{k}\) we have (even when \(p_{0}=\infty\)) \[\left\|f_{i t}\right\|_{L^{p_{0}}}=\|f\|_{L^{p}}^{\frac{p}{p_{0}}}\] since \(\left|a_{k}^{P(i t)}\right|=a_{k}^{\frac{p}{p_{0}}}\).
By the disjointness of the \(B_{j}\) ’s we have (even when \(q_{0}=1\)) \[\left\|g_{i t}\right\|_{L^{q_{0}^{\prime}}}=\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q_{0}^{\prime}}}\] since \(\left|b_{j}^{Q(i t)}\right|=b_{j}^{\frac{q^{\prime}}{q_{0}^{\prime}}}\).
Thus Hölder’s inequality and the hypothesis give \[\begin{aligned} |F(i t)| & \leq\left\|T\left(f_{i t}\right)\right\|_{L^{q_{0}}}\left\|g_{i t}\right\|_{L^{q_{0}^{\prime}}} \\ & \leq M_{0}\left\|f_{i t}\right\|_{L^{p_{0}}}\left\|g_{i t}\right\|_{L^{q_{0}^{\prime}}} \\ & =M_{0}\|f\|_{L^{p}}^{\frac{p}{p_{0}}}\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q_{0}^{\prime}}} . \end{aligned}\]
By similar calculations, which are valid even when \(p_{1}=\infty\) and \(q_{1}=1\), we have \[\left\|f_{1+i t}\right\|_{L^{p_{1}}}=\|f\|_{L^{p}}^{\frac{p}{p_{1}}},\] and \[\left\|g_{1+i t}\right\|_{L^{q_{1}^{\prime}}}=\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q_{1}^{\prime}}}.\]
Proceeding as on the previous slide we deduce that \[|F(1+i t)| \leq M_{1}\|f\|_{L^{p}}^{\frac{p}{p_{1}}}\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q^{\prime}}}.\]
We observe that \(F\) is holomorphic in the open strip \(S\) and continuous on its closure. Also, \(F\) is bounded on the closed unit strip (by some constant that depends on \(f\) and \(g\)).
Therefore, using the lemma from the previous remark, we deduce that \[\begin{aligned} |F(z)| &\leq\left(M_{0}\|f\|_{L^{p}}^{\frac{p}{p_{0}}}\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q_{0}^{\prime}}}\right)^{1-\theta}\left(M_{1}\|f\|_{L^{p}}^{\frac{p}{p_{1}}}\|g\|_{L^{q^{\prime}}}^{\frac{q^{\prime}}{q_{1}^{\prime}}}\right)^{\theta}\\ &=M_{0}^{1-\theta} M_{1}^{\theta}\|f\|_{L^{p}}\|g\|_{L^{q^{\prime}}}, \end{aligned}\] when \(\operatorname{Re} z=\theta\). Observe that \(P(\theta)=Q(\theta)=1\) and hence \[F(\theta)=\int_{Y} T(f) g d \nu\]
Taking the supremum over all finitely simple functions \(g\) on \(Y\) with \(L^{q^{\prime}}\) norm less than or equal to one, we conclude the proof of the Riesz interpolation theorem. $$\tag*{$\blacksquare$}$$
Young’s convolution inequality
If \(1\le p, q, r\le \infty\) satisfy \[\frac{1}{q}+1=\frac{1}{p}+\frac{1}{r}.\] Then for all functions \(f\in L^p(\mathbb R)\) and \(g\in L^r(\mathbb R)\) we have \[\|f*g\|_{L^q(\mathbb R)}\le \|f\|_{L^p(\mathbb R)}\|g\|_{L^r(\mathbb R)},\] where \(f*g\) is the convolution of \(f\) and \(g\), which is defined by \[f*g(x)=\int_{\mathbb R}f(x-y)g(y)dx=\int_{\mathbb R}f(y)g(x-y)dx.\]
Proof: Fix a function \(g\) in \(L^{r}(\mathbb R)\) and let \(T(f)=f * g\).
The operator \(T\) maps \(L^{1}(\mathbb R)\) to \(L^{r}(\mathbb R)\) with norm at most \(\|g\|_{L^{r}}\), (Why?).
The operator \(T\) also maps \(L^{r^{\prime}}(\mathbb R)\) to \(L^{\infty}(\mathbb R)\) with norm at most \(\|g\|_{L^{r}}\), (Why?).
The Riesz interpolation theorem gives that \(T\) maps \(L^{p}(\mathbb R)\) to \(L^{q}(\mathbb R)\) with norm at most the quantity \[\|g\|_{L^{r}}^{\theta}\|g\|_{L^{r}}^{1-\theta}=\|g\|_{L^{r}},\] where \[\frac{1}{p}=\frac{1-\theta}{1}+\frac{\theta}{r^{\prime}} \quad \text { and } \quad \frac{1}{q}=\frac{1-\theta}{r}+\frac{\theta}{\infty}.\]
This completes the proof of Young’s inequality. $$\tag*{$\blacksquare$}$$